The correct answer is option (D)

Energy of photon $=\frac{12500}{980}=12.75eV$

$\therefore$  Electron will excite to n = 4

Since 'R' $\propto$ n²

$\therefore$  Radius of atom will be 16a₀

The correct answer is option (A)

We know that $E=\frac{hc}{\lambda }$

So, for atom $E_A=\frac{hc}{{\lambda }_A}$

And for neutron $E_N=\frac{hc}{{\lambda }_N}$

Then, $\frac{E_A}{E_N}=\frac{hc}{{\lambda }_A}\times \frac{{\lambda }_N}{hc}\Rightarrow \frac{{\lambda }_N}{{\lambda }_A}$

Here, E_A is order of eV and E_N is order of MeV.

Therefore, $\frac{{\lambda }_N}{{\lambda }_A}=\frac{E_A}{E_N}=\frac{eV}{MeV}=\frac{1eV}{{10}^{6}eV}$

${\lambda }_N={10}^{-6}{\lambda }_A$

The correct answer is option (C)

We know,

Wavelength of emitted photon from n₂ state to n₁ state is

$\frac{1}{\lambda }$  =  RZ² $\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Here electron comes from n^th state to ground state (n = 1),

then the wavelength of photon is ,

$\frac{1}{{\mathrm{\Lambda }}_n}$  =  RZ² $\left(\frac{1}{1^2}-\frac{1}{n^2}\right)$

$\Rightarrow$$$ $\mathrm{\Lambda }$_n   =  $\frac{1}{R{Z}^2}{\left(1-\frac{1}{n^2}\right)}^{-1}$

As n is very large, so using binomial theorem

$\mathrm{\Lambda }$_n  =  $\frac{1}{R{Z}^2}\left(1+\frac{1}{n^2}\right)$

$\Rightarrow$$$ $\mathrm{\Lambda }$_n =  $\frac{1}{R{Z}^2}+\frac{1}{R{Z}^2}\left(\frac{1}{n^2}\right)$

We know,

$\lambda$_n =  $\frac{2\pi r}{n}$

=  2$\pi$ $\left(\frac{n^2{h}^2}{4{\pi }^{2}mZ{C}^2}\right)\times \frac{1}{n}$

$\therefore$ $\lambda$_n $\propto$ n

$\Rightarrow$$$ n  =  K $\lambda$_n

$\therefore$ $\mathrm{\Lambda }$_n  = $\frac{1}{R{Z}^2}+\frac{1}{R{Z}^2}\left(\frac{1}{{\left(K{\lambda }_n\right)}^2}\right)$

Let A  =  $\frac{1}{R{Z}^2}$  and   B  = $\frac{1}{K^{2}R{Z}^2}$

$\Rightarrow$$$ $\mathrm{\Lambda }$_n  =  A + $\frac{B}{{\lambda }_n^2}$

The correct answer is option (C)

Applying conservation of momentum :

mv + 0 = mv₁ + 2mv₂

$\Rightarrow$$$ v = v₁ + 2v₂ . . . . . (1)

As collision is elastic,

So, coefficient of restitution, e = 1

$\therefore$ e = 1 = $\frac{velocityofseparation}{Velocityofapproach}$

$\Rightarrow$$$ 1 = $\frac{v_2-v_1}{v-0}$

$\Rightarrow$$$ v = v₂ $-$ v₁ . . . . .(2)

Add (1) and (2),

2v = 3v₂

$\Rightarrow$$$ v₂ = $\frac{2v}{3}$

put value of v₂ in equation (1),

v₁ = v $-$ 2v₂

= v $-$ $\frac{4v}{3}$

= $-$ $\frac{v}{3}$

$\therefore$ Fractional loss of energy of neutron.

P_d = $\frac{k_i-k_f}{k_i}$

= $\frac{\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_1^2}{\frac{1}{2}m{v}^2}$

= $\frac{v^2-\frac{v^2}{9}}{v^2}$

= $\frac{8}{9}$

= 0.89



Applying momentum of conservation,

mv + 0 = mv₁ + 12mv₂

$\Rightarrow$$$ v = v₁ + 12v₂ . . . . . (3)

Here also e = 1

$\therefore$ e = 1 = $\frac{v_2-v{}_1}{v-0}$

$\Rightarrow$$$ v = v₂ $-$ v₁ . . . . . . (4)

adding (3) and (4), we get

2v = 13v₂

$\Rightarrow$$$ v₂ = $\frac{2v}{13}$

put this v₂ in equation (3), we get

v₁ = v $-$ 12 $\times$ $\frac{2v}{13}$

= $-$ $\frac{11v}{13}$

$\therefore$ Frictional loss

p_c = $\frac{\frac{1}{2}m{v}^2-\frac{1}{2}m{\left(\frac{11}{13}v\right)}^2}{\frac{1}{2}m{v}^2}$

= $\frac{48}{169}$

= 0.28

The correct answer is option ()

Note :

(1)   In Lyman Series, transition happens in n = 1 state
from n = 2, 3, . . . . . $\propto$

(2)   In Balmer Series, transition happens in n = 2 state
from n = 3, 4, . . . . . $\propto$

(3)   In Paschen Series, transition happens in n = 3 state
from n = 4, 5, . . . . . $\propto$

(4)   In Bracktt Series, transition happens in n = 4 state
from n = 5, 6 . . . . . . $\propto$

(5)   In Pfund Series, transition happens in n = 5 state
from n = 6, 7, . . . . $\propto$

We know,

$\frac{1}{\lambda }$  =  RZ² $\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Series limit means transition happens

from n = $\propto$ to n = 1, for Lyman Series.

In series limit for Lyman series,

$\frac{1}{{\lambda }_L}$ = RZ² $\left(\frac{1}{1^2}-\frac{1}{\propto }\right)$

$\Rightarrow$$$ $\frac{1}{{\lambda }_L}$ = RZ²

We know,

E = $\frac{hc}{\lambda }$ = h$\gamma$

$\Rightarrow$$$ $\gamma$ = $\frac{c}{\lambda }$

So, frequency in Lyman Series,

$\gamma$_L =  $\frac{c}{{\lambda }_L}$  = c $\times$ RZ²

In Pfund series,

n₂ = $\propto$  and  n₁ = 5

$\therefore$ $\frac{1}{{\lambda }_P}$ = RZ²$\left(\frac{1}{5^2}-\frac{1}{{\propto }^2}\right)$

$\Rightarrow$$$ $\frac{1}{{\lambda }_P}$ = $\frac{R{Z}^2}{25}$

$\therefore$ ${\gamma }_P$   =   $\frac{c}{{\lambda }_P}$ = c $\times$ $\frac{R{Z}^2}{25}$

$\therefore$ $\gamma$_P = $\frac{cRZ{}^2}{25}$ = $\frac{{\gamma }_L}{25}$    [as   $\gamma$_L = cRZ²]