JEE PYQ: Atomic Physics Question 53
Question 53 - 2019 (10 Apr Shift 2)
In Li$^{++}$, electron in first Bohr orbit is excited to a level by a radiation of wavelength $\lambda$. When the ion gets deexcited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of $\lambda$?
(Given: $h = 6.63 \times 10^{-34}$ Js; $c = 3 \times 10^8$ ms$^{-1}$)
(1) 11.4 nm
(2) 9.4 nm
(3) 12.3 nm
(4) 10.8 nm
Show Answer
Answer: (4)
Solution
Spectral lines obtained on account of transition from $n^{th}$ orbit to various lower orbits is $\frac{n(n-1)}{2}$. $6 = \frac{n(n-1)}{2} \Rightarrow n = 4$. $\Delta E = \frac{Z^2}{n^2} \times 13.6$ eV. $\frac{1}{\lambda} = Z^2\left(\frac{13.6 \text{eV}}{hc}\right)\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$. $= (13.4)(3)^2\left[1 - \frac{1}{16}\right]$. $\lambda = \frac{1242 \times 16}{(13.4) \times (9)(15)}$ nm $= 10.8$ nm.
BETA
