Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

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Answer- (a)

Explanation-The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because of the electrons not being subject to a central force.

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Answer- c

Explanation- when one decides to work in a frame of reference where the electron is at rest, the given expression is not true as it the non-inertial frame of reference.

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Answer- (c)

Explanation- The atomic number of lithium is 3, therefore, the radius of Li+ ion in its ground state, on the basis of Bohr's model, will be about 1/3 times to that of Bohr radius.

Therefore, the radius of lithium ion is near 53/3=18≈ pm.

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Explanation- Hγ in Balmer series corresponds to transition n= 5 to n= 2. So, the electron in ground state i.e., from n= 1must first be placed in state n= 5.

Energy required for the transition from n= 2 to n= 5 is given by E₁-E₅=13.6-0.54=13.06eV

Since, angular momentum is conserved,

angular momentum corresponding to Hg photon= change in angular momentum of electron

so L₅-L₂=5h-2h= 3h = 3 $*1.06*{10}^{-34}$

=3.18 $*{10}^{-34}$ kgm²/s

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Explanation- The frequency of any line in a series in the spectrum of hydrogen like atoms corresponding to the transition of electrons from (n+ p) level to nth level can be expressed as a difference of two terms;

From the formula we can say that v=cRZ² { $\frac{1}{({n+p)}^2}-\frac{1}{n^2}$ }

Where p =1,2,3, … R os rydberg constant

When p<

V=cRZ² { $\frac{1}{n^2}({1+\frac{p}{n})}^{-2}-\frac{1}{n^2}$ }

V=cRZ² { $\frac{1}{n^2}-\frac{2p}{n^3}-\frac{1}{n^2}\}$

By binomial theorem

V=cRZ² { $\frac{2p}{n^3}=\frac{2cR{Z}^2}{n^3}p$

Thus, the first few frequencies of light that is emitted when electrons fall to the nth level

from levels higher than n, are approximate harmonic (i.e., in the ratio 1 : 2 : 3 …) when

n >> 1 .

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Explanation- The electron in Hydrogen atom in ground state revolves on a circular path whose radius is equal to the Bohr radius (a_n) . Let the velocity of electron is v

∴ Number of revolutions per unit time= $\frac{2\pi a_0}{V}$

The electric current is given by i=q/t, if q charge flows in time t. Here,   q=e

The electric current is given by i= $\frac{2\pi a_0}{V}e$

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Explanation- For a He -nucleus with charge 2e and electrons of charge− e, the energy level in ground state is

E=-13.6Z²/n²= -54.4eV

Thus, the ground state will have two electrons each of energy E and the total ground state

energy would be   - (4*13.6)eV = -54.4eV.

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Explanation- The total energy of the electron in the stationary states of the hydrogen atom is given by E_n = - $\frac{m{e}^4}{8n^2{\epsilon }_0^2{h}^2}$

where signs are as usual and the m that occurs in the Bohr formula is the reduced mass of

electron and proton. Also, the total energy of the electron in the ground state of the

hydrogen atom is−13.6 eV. For H-atom reduced mass m_e . Whereas for positronium, the

reduced mass is m=m_e/2

Hence, the total energy of the electron in the ground state of the positronium atom is

 -13.6/2=-6.8eV

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Explanation- According to Bohr model electrons having different energies belong to different levels having different values of n. So, their angular momenta will be different, as

L=nh/2 $\pi L\propto$ n

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Explanation- If proton had a charge (+4/3)e and electron a charge (-3/4)e, then the Bohr formula for the H-atom remain same, since the Bohr formula involves only the product of the charges which remain constant for given values of charges.