JEE Advanced 2023 Paper 2 · Q11 · Atomic Structure

Question

For $He^+$, a transition takes place from the orbit of radius $105.8$ pm to the orbit of radius $26.45$ pm. The wavelength (in nm) of the emitted photon during the transition is ___. [Use: Bohr radius $a_0 = 52.9$ pm; Rydberg constant $R_H = 2.2	imes 10^{-18}$ J; $h = 6.6	imes 10^{-34}$ J·s; $c = 3	imes 10^8$ m·s$^{-1}$]

SolveKar AI · Accepted Answer

For $He^+$ ($Z = 2$), $r_n = n^2 a_0/Z$. With $a_0/Z = 52.9/2 = 26.45$ pm: $r_1 = 26.45$ pm ($n=1$), $r_2 = 4\cdot 26.45 = 105.8$ pm ($n=2$). Transition $n=2 	o n=1$: $\Delta E = R_H Z^2 (1/1 - 1/4) = 2.2	imes 10^{-18}\cdot 4\cdot 3/4 = 6.6	imes 10^{-18}$ J. Wavelength $\lambda = hc/\Delta E = (6.6	imes 10^{-34})(3	imes 10^8)/(6.6	imes 10^{-18}) = 3	imes 10^{-8}$ m = $30$ nm.

JEE Advanced 2023 Paper 2 · Q11 · Atomic Structure

Solution