NCERT Solutions for Class 11 Science Chemistry Chapter 2 - Structure Of Atom

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Answer:

Rutherford’s α-particle scattering experiment gave the following conclusions:
1- Most of the space in the atom is empty as most of the alpha particle passed undeflected through the atom.
2- The radius of the atom is about 10^–10 m while that of nucleus is 10^–15 m , this signifies that size of atom is 10⁵ times more than size of nucleus.
3- Electrons and the nucleus are held together by electrostatic forces of attraction.

But, the conclusion that electrons move in a circular path of fixed energy called orbits was given by Bohr and not Rutherford.

Hence, the correct answer is option C.


 

Answer:

The ground state electronic configuration along with their corresponding elements are mentioned below:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s² - Ground state electronic configuration of Ni.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹- Ground state electronic configuration of Cu.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹- Ground state electronic configuration of Cr.

Therefore, 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s² is not the ground state electronic configuration of Cu due to symmetrical electronic configuration and electron exchange energy Cu exist in 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ ground state electronic configuration.

Hence, the correct answer is option B.

Answer:

s-orbitals are spherical in shape, the probability of finding electron is maximum near the nucleus and the probability of finding the electron at a given distance is equal in all directions as the number of angular nodes are zero (l=0).

Answer:

Following are the characteristics of cathode rays :
1- Cathode rays are the rays which contain negatively charged particle hence they move from cathode (negative terminal) to anode (positive terminal)
2-In the absence of an external electric or magnetic field these rays travel in straight line while in presence of external electric and magnetic field these rays behave in the similar manner as negatively charged particle would behave.
3- Characteristics of cathode rays do not depend upon the material of electrodes in cathode ray tube and upon the nature of gas present in the cathode ray tube as cathode ray always consist of electrons, no matter what is the nature of electrode used or what is the nature of gas used.

As, characteristics of cathode rays do not depend upon the nature of gas present in the cathode ray tube.

Hence, the correct answer is option D.

Answer:

The following are the characteristics of an electron:
1-An electron is a negatively charged particle.
2-The mass of electron is $9.1\times {10}^{-31}$ kg which is very small in comparison to the mass of neutron.
3-An electron is basic constituent of all the atoms.
4-It is a constituent of cathode rays.

Hence, the correct answer is option B.

Answer:

Thomson proposed a plum pudding model of atom. According to Thomson Model, an atom is a uniform sphere of positively charged particle and equal number of electrons are embedded on this positively charged sphere. Hence, overall neutrality of atom was explained by Thomson. 
While spectra of hydrogen atom, position of electrons, protons and neutrons in atom and stability of atom was explained by Bohr Model of atom.

Hence, the correct answer is option A.

Answer:

Two atoms are said to be isobars if they have same mass number but different atomic number or if sum of the number of protons and neutrons is same but the number of protons is different.

If they have same atomic number but different mass number they are called isotopes.

Hence, the correct answer is option D.

Answer:

Number of radial nodes is given by the formula (n-l-1)  where  n is principal quantum number and l is azimuthal quantum number.

For 3p orbital n is 3 and  l is 1. So, the number of radial nodes is 3-1-1 that is equal to 1.

Hence, the correct answer is option D.
 

Answer:

Number of angular nodes is given by the value of l (azimuthal quantum number). 

As the value of l for 4d is 2 so, number of angular nodes for 4d orbital would be given as 2.

Hence, the correct answer is option C.

Answer:

Heisenberg’s uncertainty principle states that the position and velocity of an electron cannot both be measured simultaneously with 100% accuracy. Hence, it is responsible to rule out the existence of definite paths or trajectories of an electron, as it is not possible to know the exact path of an electron at a given time.

Hence the correct answer is option B.

Answer:

Total number of orbitals associated with a given shell is given by the formula n² , where n is principal quantum number. Hence the total number of orbitals in 3rd shell would be 3² that is 9. These orbitals would be present in sub shell 3s, 3p and 3d. 3s contain 1 orbital, 3p contain 3 orbitals, 3d contains 5 orbitals. Therefore, in total 9 orbitals are present.

Hence the correct answer is option C.

Answer:

Orbital angular momentum is given by the formula $\sqrt{l(l+1)}$ $\frac{h}{4\pi }$ . Hence the value of orbital angular momentum depends on the value of l. 
Hence, the correct answer is option A.

Answer:

Let, the percentage abundance of Cl-37 be A

Let, the percentage abundance of Cl-35 be B

Let, the molar mass of Cl-37 be M₁ = 37

Let, the molar mass of Cl-35 be M₂ = 35

A + B = 100

B = 100 - A

M_avg = $\frac{M_1\times A+M_2\times B}{A+B}$

35.5 = $\frac{37\times A+35(100-A)}{100}$

35.5 = $\frac{\left(37\times A\right)+3500-\left(35\times A\right)}{100}$

Hence, A = 25 

B = 100 - 25 = 75 

$\frac{A}{B}= \frac{25}{75}=\frac{1}{3}$

Thus, the ratio of Cl-37 and Cl-35 is approximately 1:3.

Hence, the correct answer is option C.

Answer:

The electronic configuration of Cr = [Ar] 3d⁵ 4s¹

                                                Cr³⁺ = [Ar] 3d³ 4s⁰

the electronic configuration of Fe =  [Ar] 3d⁶ 4s²

                                               Fe³⁺ = [Ar] 3d⁵ 4s⁰

the electronic configuration of Mn = [Ar] 3d⁵ 4s²

                                              Mn²⁺ = [Ar] 3d⁵ 4s⁰

the electronic configuration of Co = [Ar] 3d⁷ 4s²

                                              Co³⁺ = [Ar] 3d⁶ 4s⁰

the electronic configuration of Sc = [Ar] 3d¹ 4s²

                                              Sc³⁺ = [Ar] 3d⁰ 4s⁰ [noble gas configuration]

Thus, electronic configuration of Fe³⁺ is same as Mn²⁺ that is [Ar] 3d⁵ 4s⁰.

Hence, the correct answer is option B. 
 

Answer:

According pauli exclusion principle, no two electrons present in the same orbital can have same set of all the four quantum numbers. So, the correct statement is that, two electrons present in 2s orbital have same spin quantum numbers m_s but of opposite sign according to pauli exclusion principle.

Hence, the correct answer is option D.

Answer:

We know, λ =$\frac{h}{mv}$. For same value of v (velocity), larger the value of m (mass), shorter is the λ (wavelength).
Since, alpha particles have the largest mass, they have the shortest wavelength.

Hence, the correct answer is option B.

Answer:

Isotopes are species of same elements having different mass number (no. of protons + no. of neutrons) , but they have the same atomic number (number of protons). Therefore, the entities (iii) and (iv) are not isotopes as they have different atomic numbers.

Options (i) and (ii)  are definitely isotopes.

Hence, the correct answers are options C and D.

Answer:

Degenerate orbital means orbital having same energies. And the energy of an electron in a multi-electron system of atoms depends on the value of principal quantum number (n) and the azimuthal quantum number ( l) as well. If the value of n and l is same, that means the orbitals are degenerate.

(i) (a) n = 3, l = 2, m_l = –2, m_s =  and (b) n = 3, l = 2, m_l = –1, m_s= −
(iv) (a) n = 3, l = 2, m_l = +2, m_s = − and (b) n = 3, l = 2, m_l = +2, m_s = + 

have n and l values as equal so they are degenerate.

Answer:

If n = 1, l ≠ 1. Hence, (i) is wrong.
If n = 2, l= 0, 1. For l = 1, m = -1, 0, +1. Hence (ii) is correct.
If n = 3, l= 0, 1, 2. For l = 2, m = -2, -1, 0, +1, +2. Hence (iii) is correct.
If n = 3, l ≠4. Hence, (iv) is wrong.

Hence, the correct answers are options B and C.

Answer:

Na+, Mg²⁺ have 10 electrons.

Na+, O^2– have 10 electrons

Answer:

Azimuthal quantum number or l gives the value of orbital angular momentum of a certain orbital. And it defines the 3-D shape of an orbital by introducing the concept of angular momentum.
Spin quantum number m_s gives the value of spin angular momentum of an electron as electron rotates around its axis, therefore besides having charge and mass an electron also has a certain orientation of the spin relative to the chosen axis which gives rise to the concept of spin angular quantum number (m_s).
Hence statements (i) and (iv) are correct.

Hence, the correct answers are options A and D.

Answer:

Z_eff experienced by s sub-shell is maximum while d sub-shell is minimum , this is because s subshell is spherically symmetrical therefore it has maximum penetration power and hence, it remains nearer to the nucleus and experience maximum Z_eff. While d subshell have double dumbbell shape hence, it is more diffused and therefore have minimum penetration power, it stays away from the nucleus and experiences minimum  Z_eff.
Hence, the value of Z_eff  in increasing order is given as $d<p<s$.
 

Answer:

Oxygen atomic number is 8.
Electronic configuration is 1s² 2s² 2p⁴.
Distribution of electrons in orbitals is as follows:

 

Answer:

Atomic number of nickel is 28 and its electronic configuration is as follows:

In , there will be two electrons less than the number of electrons present in a neutral nickel atom.
Therefore, the electronic configuration is as follows:

Hence, Nickel loses two electrons from 4s orbital as the 4th shell is the outermost shell of the atom
       

 

Answer:

Degenerate orbitals are the one, which have same values of n and l, that is same values of principal and azimuthal quantum numbers.
Hence, $3d_{xy }3d_{z^2}$ are degenerate orbitals and also $4d_{xy}4d_{yz}4d_{z^2}$ are degenerate orbitals.
 

Answer:

For 3p orbital, n = 3, l = 1
Number of angular nodes = l =1
Number of radial nodes =$n - l -1$ = $3-1-1$ = 1

Hence, number of angular node is 1 and number of radial nodes is also 1.

Answer:

I (a) (n+l) values are for 1s = 1 + 0 = 1, 2s = 2 + 0 = 2, 3s = 3 + 0 = 3, 2p = 2 + 1 = 3
Hence, increasing order of their energy is $1s<2s<2p<3s$.

(b) 4s = 4 + 0 = 4, 3s = 3 + 0 = 3, 3p = 3 + 1 = 4, 4d = 4 + 2 = 6. Hence, $3s<3p<4s<4d.$

(c) 5p = 5 + 1 = 6, 4d = 4 + 2 = 6, 5d = 5 + 2 = 7, 4f = 4 + 3 = 7, 6s = 6 + 0 = 6. Hence, $4d<5p<6s<4f<5d$.

(d) 5f = 5 + 3 = 8, 6d = 6 + 2 = 8, 7s = 7 + 0 = 7, 7p = 7 + 1 = 8. Hence, $7s<5f<6d<7p$.

II. (a) (n+l) values are for 4d = 4 + 2 = 6, 4f = 4 + 3 = 7, 5s = 5 + 0 = 5, 7p = 7+1 =8 . Hence, 5s has the lowest energy. 

     (b) 5p = 5 + 1 = 6, 5d = 5 + 2 = 7, 5f = 5 + 3 = 8, 6s = 6 + 0 = 6, 6p = 6 + 1 = 7. Hence, 5f has highest energy.
      

 

Answer:

Charged particles can be disturbed by the presence of either electric or magnetic field.

Proton is a positively charged particle - so it will have deflection.

Cathode ray contains electrons that is negatively charged particle - so it will have deflection.

Electrons are negatively charged particle - so it will have deflection.

Neutron is electrically neutral - so will not have any deflection on the path.

Hence, Neutron will not have any deflection from the path on passing through an electric field.

Answer:

Atomic number of the atom = mass number - number of neutrons

As, mass number is given as 13 and number of neutrons is given is 7.

Therefore,  Atomic number = 13 - 7 = 6

Hence, the atomic number of the atom is 6.

Answer:

λ(A) = 300 nm, $300nm=300\times {10}^{-9}m=3\times {10}^{-7}m$

λ(B) = 300 μm, $300\mu m=300\times {10}^{-6}m=3\times {10}^{-4}m$

λ(C) = 3 nm, $3nm = 3\times {10}^{-9}m$

λ(D) = 30 A°, $30\AA =30\times {10}^{-10}m=3\times {10}^{-9}m$

As, $E=hv=\frac{hc}{\lambda }$ So, $E\propto \frac{1}{\lambda }$.
 Hence, increasing order of energy is $B<A<C=D$.<A<C=D

 

Answer:

Configuration which have either half-filled or fully filled orbitals are more stable due to symmetrical distribution of electrons and maximum exchange energy in $3d^{10}4s^1$ d-orbitals are completely filled and s-orbital is half-filled. Hence it is more stable configuration.

Answer:

$$\begin{gathered} \stackrel{\mathrm{\_}}{\mathrm{\nu }=}{\mathrm{R}}_{\mathrm{H}}\left(\frac{1}{{{\mathrm{n}}_1}^2}-\frac{1}{{{\mathrm{n}}_2}^2}\right) {\mathrm{cm}}^{-1} \\ = 109677\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \\ = 20564.44 {\mathrm{cm}}^{-1} \end{gathered}$$

Answer:

Given mass of ball (m) = 100 g = 0.1 kg, given velocity of ball (v) = 100 km/h = $\frac{100 \times 1000 m}{ 60 \times 60 s}$ = $\frac{1000 m}{36 s}$
λ=$\frac{h}{mv}$=$\frac{6.626\times {10}^{-34}kg{m}^2{s}^{-1}}{0.1kg\times 1000/36m{s}^{-1}}$=$238.5\times {10}^{-36}$ m

As the wavelength of ball is very small, the wave nature cannot be detected.

Answer:

The bright line spectrum shows that the energy levels in an atom are quantized. These lines corresponds to definite wavelenghts and are obtained as a result of electronic transitions between the energy levels.
Hence, the electrons in these levels have quantized values.

Answer:

The de-broglie equation states,  ${}_{v=\frac{h}{m\lambda }}$

The velocity here is inversely proportional on the masses.
Higher the mass lower will be the velocity. As, electron has lower mass than proton.
Hence, electron will have a higher velocity to produce matter waves of same wavelength.

Answer:

Wavelength is defined as the distance between two successive peaks and two successive troughs of the wave.

Hence, the wavelength of the radiation would be $\lambda =$$4\times 2.16 pm$ = 8.64 pm.=0.6494×10−6m 

Answer:

Wavelength of light is given by the formula $\lambda =\frac{h}{mv}$.

Given, frequency $\nu =4.620\times {10}^{14} Hz=4.620\times {10}^{14} s^{-1}$

$\lambda =\frac{c}{\nu }=\frac{3\times {10}^{18} m{s}^{-1}}{4.620\times {10}^{14} s^{-1}}$ = 0.6494 × 10^-6 m = 649.4 nm

Hence, wavelength of radiation in nanometer is 649.4 nm it belongs to visible region.
     

Answer:

Answer:

Given that, speed = 90 m/s
Mass = 10 g = 10 × 10 ⁻³ kg

Uncertainly in speed $\left(\Delta v\right)=\frac{4\times 90}{100}$ = 36 m/s
From Heisenberg uncertainty principle $\Delta x.\Delta v\ge \frac{h}{4\pi m}$  
$\Delta x=\frac{h}{4\pi m\Delta v}$ = $\frac{6.626\times {10}^{-34} kg{m}^2{s}^{-1}}{4\times 3.14\times 10\times {10}^{-3}kg\times 3.6 m{s}^{-1}}$  = $1.46\times {10}^{-33} m$

 Hence, uncertainty in speed and position wold be 36 m/s and $1.46\times {10}^{-33} m$ respectively.

Answer:

From Heisenberg uncertainty principle $\Delta x.\Delta v\ge \frac{h}{4\pi m}$  as the uncertainty is in inverse proportion with the mass we can say that greater mass will bring smaller in insignificant uncertainty for macroscopic particles.

If uncertainty principle is applied to an object of mass, say about a milligram (10⁻⁶ kg) then 

$\Delta x.\Delta v=\frac{h}{4\pi m}$ = $\frac{6.626\times {10}^{-34} kg{m}^2{s}^{-1}}{4\times 3.14\times {10}^{-6}kg}$  = $0.52 \times {10}^{-28} m^2{s}^{-1}$

The value of  obtained is extremely small and is insignificant.

Therefore, for miligram-sized or heavier objects, the associated uncertainties are hardly of any real consequence.
 

Answer:

In case of hydrogen atom, the energies of the electron in different orbitals depends only on the value of n. Hence, different orbitals of the same shell have same energy. However, in case of multi electron atoms, the energies of the orbitals depend upon n+l values. Hence, for the same value of n but different value of l, i.e., different sub shells belonging to the same main shell have different energies.

Answer:

The electronic configuration of the respective ions/element may be given as follows :

 (i) Electronic configuration of Cu =  2,8,18,1 i.e. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹â° 4s¹

(ii) Electronic configuration of Cu²âº = 2,8,17 i.e. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹

(iii) Electronic Configuration of Zn²âº = 2,8,18,2 i.e. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹â°

(iv) Electronic Configuration of Cr³âº = 2,8,8,3 i.e. 1s² 2s² 2p⁶ 3s² 3p⁶ 4d³

Hence, the correct matching of the electronic configuration is

 (i) $\stackrel{}{\to }$ (c) ; (ii) $\stackrel{}{\to }$ (d) ; (iii) $\stackrel{}{\to }$ (a) ; (iv) $\stackrel{}{\to }$ (e).

 

Answer:

The correct match of quantum number with the provided information can be represented as follows:
 

Answer:

The correct match of rules with the provided statements can be represented as follows:
 

Answer:

The correct match of Rays with the provided frequencies can be represented as follows:
 

Answer:

The correct match can be represented as follows:
 

Answer:

The electronic configuration of the respective ions/element may be given as follows:
 

Answer:

Both assertion and reason are true and reason is the correct explanation of assertion.
Isotopes have the same atomic number i.e. same number of electrons. Electrons are responsible for chemical behaviour of any element.
Therefore, isotopes exhibit similar chemical properties.
Hence, the correct answer is option A.

Answer:

Both A and R are true and R is not the correct explanation of A .This can be explained as follows:
A body which absorbs and emits all radiations falling on it is called perfect black body. As temperature $\alpha$ frequency, with rise in temperature, frequency of radiation emitted by a body increases.

Hence, the correct answer is option B.

Answer:

Assertion is true and reason is false. 
According to Heisenberg's uncertainly principle the exact position and exact momentum of an electron cannot be determined simultaneously. Thus, the path of electron in an atom is not clearly defined.

Hence, the correct answer is option C.

Answer:

When radiation with certain minimum frequency $v_0$ strike the surface of a metal, electrons are ejected from the surface of the metal this phenomenon is called photoelectric effect. 
 
Experiment showing photoelectric effect:

Light of a particular frequency strikes a clean metal surface inside a vacuum chamber. Electrons are ejected from the metal and are counted by a detector that measures their kinetic energy.

The result observed in this experiment were:
 
(i) The electrons are ejected from the metal surface as soon as the beam of light strikes the surface. i.e. there is no time lag between the striking of light beam and the ejection of electrons from the metal surface. 
(ii) The number of electrons ejected is proportional to the intensity or brightness of light. 
(iii) For each metal, there is a characteristic minimum frequency, $v_0$ (also known as threshold frequency) below which photoelectric effect is not observed. At a frequency $v>$$v_0$ the ejected electrons come out with certain kinetic energy. The kinetic energies of these ejected electrons increases with the increase of frequency of the light used.

The above observation cannot be explained by the electromagnetic wave theory. According to this theory, since radiations were continuous therefore, it should be possible to accumulate energy on the surface of the metal irrespective of it frequency and thus, radiations of all frequencies should be able to eject electrons. 
Similarly according to this theory the energy of the electrons ejected should depend upon the intensity of the incident radiation. 

Particle Nature of Electromagnetic Radiation 
To explain the phenomena of black body radiation and photoelectric effect Max planck in 1900, put forward a theory known after his name as Planck's quantum theory. This theory was further extended by Einstein in 1905. 

(i) The radiant energy is emitted or absorbed not continuously but discontinuously in the form of small discrete packets of energy each such packet of energy is called a quantum, in case of light quantum of energy is called a photon. 
(ii) The energy of each quantum is directly proportional to the frequency of the radiation 
i.e. $E \alpha \nu$ or E = $hv$.
where h is a proportionality constant called planck's constant. Its value to approximately equal to 6.626×10⁻³⁴ Js. 
(iii) The total amount of energy emitted or absorbed by a body will be some whole number quanta. 
Hence,E = n$hv$ (where, n is any integer).

Answer:

We know, $h\nu = h{v}_0 + K.E.$

$v$ = 1.0 × 10¹⁵ s^–1 , K.E. = 1.988 × 10^–19 J , ${\displaystyle \mathrm{h }= 6.626\times {10}^{^-\mathrm{³⁴ }}Js}$ 

$v_0$ = $1.0 \times {10}^{15} s^{-1} - \frac{1.988 \times {10}^{-19} J}{ 6.626 \times {10}^{-34} Js}$ = $7.0\times {10}^{14} s^{-1} Hz$

Frequency of striking photon may be calculated as follows : 

$\nu =\frac{c}{\lambda }$ = $\frac{3.0\times {10}^8 m{s}^{-1}}{600\times {10}^{-9} m}$ = $5.0\times {10}^{14} s^{-1} Hz$

Since the frequency of the striking photon is less than the threshold frequency, an electron will, not be emitted from the metal surface under the influence of the influence of the striking photon.


 

Answer:

Two important points of Bohr's model that can be used to derive the given formula are: 
(i) Electrons revolve around the nucleus in stationary states i.e. orbits with fixed values of energy (quantized values). 
(ii) Energy is emitted or absorbed only when electron jumps from one orbit to another.

Derivation:
Step (i) The energy of the electron in the nth stationary state is given by : $E_n=- \frac{2{\pi }^{2}m{e}^4}{n^2{h}^2}$
where m = mass of electron, e = charge on electron, h = Planck's constant.

Step (ii) When the electron jumps from outer stationary state n₂ to inner stationary state n₁ the different of energy (ΔE) is emitted i.e.$\Delta E=E_2-E_1$ = $\frac{2{\pi }^{2}m{e}^4}{h^2} (\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})$ 

​Step (iii), ΔE = $h\nu$ = $\frac{hc}{\lambda }$ 

$\frac{1}{\lambda }=\frac{\Delta E}{hc}=\frac{2{\pi }^{2}m{e}^4}{ch3} (\frac{1}{{n_1}^2} - \frac{1}{{n_2}^2})$

Substituting the value of the constant, π,m,e in C.G.S units, we get:
$\frac{1}{\lambda } = 109677 c{m}^{-1 } (\frac{1}{{n_1}^2} - \frac{1}{{n_2}^2})$

$\frac{1}{\lambda }=\overline{v}$  which is also called as wave number 

Therefore, $\overline{v}$ = $\frac{1}{\lambda } = 109677 c{m}^{-1 } (\frac{1}{{n_1}^2} - \frac{1}{{n_2}^2})$
 

Answer:

In hydrogen spectrum the spectral lines are expressed in term of wave number $\frac{1}{\lambda } =\overline{ v}$ obey the following formula :

$\overline{v }= \frac{1}{\lambda } = 109677 c{m}^{-1 } (\frac{1}{{n_1}^2} - \frac{1}{{n_2}^2})$ 

$\overline{v } = 109677 c{m}^{-1 } (\frac{1}{2^2} - \frac{1}{3^2})$ = $109677 \times \frac{5}{36}$ = $15232 c{m}^{-1}$

$\lambda = 6.564 \times {10}^{-5 }cm$ = $6.564\times {10}^{-7}m$ 

E = $\frac{hc}{\lambda }$ = $\frac{6.626 \times {10}^{-34} Js \times 30 \times {10}^8 m{s}^{-1}}{6.564 \times {10}^{-7} m}$ = 3.028 × 10⁻¹⁹ J

$\nu = \frac{c}{\lambda }$ = $\frac{3.0 \times {10}^8 m{s}^{-1} }{6.564 \times {10}^{-7}m }$ = $0.457 \times {10}^{15 }{s}^{-1}$

Hence, the energy is 3.028 × 10⁻¹⁹ J and frequency is  $0.457 \times {10}^{15 }{s}^{-1}$.

Answer:

Bohr Model atom suffered from certain limitations :

1) According to Bohr, the radiation results when an electron jumps from one energy orbit to another energy orbit, but how this radiation occurs is not explained by Bohr.
2) Bohr Theory had explained the existence of various lines in H-spectrum, but it predicted that only a series of lines exist. At that time this was exactly what had been observed. However, as better instruments and techniques were developed, it was realized that the spectral line that had been thought to be a single line was actually a collection of several lines very close together (known as fine spectrum). 
3) Thus the appearance of the several lines implies that there are several energy levels, which are close together for each quantum number   n. This would require the existence of new quantum numbers.
4) Bohr's theory has successfully explained the observed spectra for hydrogen atom and hydrogen like ions (e.g. He⁺, Li²⁺, Be³⁺ etc), it can not explain the spectral series for the atoms having a large number of electrons.
5) There was no satisfactory justification for the assumption that the electron can rotate only in those orbits in which the angular momentum of the electron (mvr) is a whole number multiple of h/2$\mathrm{\pi }$, i.e. he could not give any explanation for using the principle of quantisation of angular momentum and it was introduced by him arbitrarily.
6) Bohr assumes that an electron in an atom is located at a definite distance from the nucleus and is revolving round it with definite velocity, i.e. it is associated with a fixed value of momentum. This is against the Heisenberg's Uncertainty Principle according to which it is impossible to determine simultaneously with certainty the position and the momentum of a particle.
7) No explanation for Zeeman effect: If a substance which gives a line emission spectrum, is placed in a magnetic field, the lines of the spectrum get split up into a number of closely spaced lines. This phenomenon is known as Zeeman effect. Bohr's theory has no explanation for this effect.
8) No explanation of the Stark effect: If a substance which gives a line emission spectrum is placed in an external electric field, its lines get split into a number of closely spaced lines. This phenomenon is known as Stark effect. Bohr's theory is not able to explain this observation as well.

The concept of the movement of an electron in an orbit was no longer considered to be proper. The concept of probability of finding the electron in an atom was based upon Heiseberg's Uncertainty principle. The modified model of atom is known as quantum mechanical  model.