A free hydrogen atom after absorbing a photon of wavelength $λ_{a}$ gets excited from the state n = 1 to the state n = 4. Immediately after that the electron jumps to n=m state by emitting a photon of wavelength $λ_{e}$ . Let the change in momentum of atom due to the absorption and the emission are $Δ p_{a}$ and $Δ p_{e}$ , respectively. If $λ_{a} / λ_{e} = \frac{1}{5}$ , which of the option (s) is /are correct ?
[Use $h c = 1242 eVnm; 1 nm = 10^{- 9} m, h$ , and c are Planck’s constant and speed of light, respectively]

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$Δ p_{a} / Δ p_{e} = \frac{1}{2}$

The ratio of kinetic energy of the electron in the state n = m to the state n = 1 is $\frac{1}{4}$

m = 2

$λ_{e}$ = 418 nm

answer is B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \frac{1}{λ_{a}} = R (1 - \frac{1}{16}) \\ \frac{1}{λ_{e}} = R (\frac{1}{m^{2}} - \frac{1}{4^{2}}) \\ Given : \frac{λ_{a}}{λ_{e}} = \frac{(\frac{1}{m^{2}} - \frac{1}{4^{2}})}{\frac{15}{16}} = \frac{1}{5} \\ \Rightarrow \frac{1}{m^{2}} = \frac{1}{4^{2}} + \frac{3}{16} \\ \Rightarrow \frac{1}{m^{2}} = \frac{1}{4} \Rightarrow m = 2 \\ Now, we know that Kinetic energy \propto \frac{1}{n^{2}} \\ \frac{K_{m}}{K_{1}} = \frac{1}{2^{2}} \times 1 = \frac{1}{4} \\ Now, 13.6 (\frac{1}{4} - \frac{1}{16}) = \frac{1242}{λ_{e}} \\ \Rightarrow 13.6 (\frac{3}{16}) = \frac{1242}{λ_{e}} \\ \Rightarrow λ_{e} = 487 nm \end{array}$