The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A…

Q: The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A° . The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

(A) 1215 A°

(B) 1640 A°

(C) 2430 A°

(D) 4687 A°

Ans: (A)

Sol: $\large \frac{\lambda_1}{\lambda_2} = \frac{Z_1^2 (\frac{1}{n_f^2} – \frac{1}{n_i^2})_1}{Z_2^2 (\frac{1}{n_f^2} – \frac{1}{n_i^2})_2} $

$\large \lambda_2 = \frac{\lambda_1 Z_1^2 (\frac{1}{n_f^2} – \frac{1}{n_i^2})_1}{Z_2^2 (\frac{1}{n_f^2} – \frac{1}{n_i^2})_2} $

On Putting the given values ,

$\large \lambda_2 = \frac{(6561 A^o) ( 1)^2 (\frac{1}{2^2} – \frac{1}{3^2})_1}{(2)^2 (\frac{1}{2^2} – \frac{1}{4^2})_2} $

= 1215 A°

A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV…

Q: A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?

(A) 2 photons of energy 10.2 eV

(B) 2 photons of energy 1.4 eV

(C) One photon of energy 10.2 eV and an electron of energy 1.4 eV

(D) One photon of energy 10.2 eV and another photon of energy 1.4 eV

Ans: (C)

Sol: The first Photon will excite the hydrogen atom in ground state to first excited state as $E_2 -E_1 = 10.2 eV$ . Hence during de-excitation a photon of 10.2 eV will be released .

The second photon of energy 15 eV  can ionise the atom . Hence balance Energy = 15-13.6 = 1.4 eV is retained by the electron . Therefore by second photon an electron of energy 1.4 eV will be released .

If the atom 100Fm257 follows the Bohr’s model and the radius of last orbit of 100Fm257 is n times the Bohr radius

Q: If the atom 100Fm257 follows the Bohr’s model and the radius of last orbit of 100Fm257  is n times the Bohr radius, then find n

(A) 100

(B) 200

(C) 4

(D) 1/4

Ans: (D)

Sol: $\large r_m = 0.53 \frac{m^2}{Z} A^o$

$\large r_m = n \times 0.53 A^o$

$\large \frac{m^2}{Z} = n $

m = 5 for 100Fm257  (the outermost shell) and Z = 100

$\large n = \frac{5^2}{100} = \frac{1}{4} $

The electric potential between a proton and an electron is given by $V = V_0 ln \frac{r}{r_0}$ , where r0 is a constant.

Q: The electric potential between a proton and an electron is given by $V = V_0 ln \frac{r}{r_0}$ , where r0 is a constant. Assuming Bohr’s model to the applicable, write variation of rn with n, n being the principal quantum number

(A) rn  ∝ n

(B) rn ∝ 1/n

(C) rn ∝ n2

(D) rn ∝ 1/n2

Ans: (A)

Sol: $\large U = e V = e V_0 ln \frac{r}{r_0}$

$\large F = |-\frac{dU}{dr}| = \frac{e V_0}{r}$

This force will provide necessary centripetal force.

$\large \frac{m v^2}{r} = \frac{e V_0}{r} $

$\large v = \sqrt{\frac{e V_0}{m}}$  …(i)

According to Bohr’s theory;

$\large m v r = n\frac{h}{2\pi}$ …(ii)

Dividing (ii) by (i)

$\large m r = ( \frac{n h}{2\pi}) \sqrt{\frac{m}{e V_0}}$

rn  ∝ n