Physics - Modern Physics Question with Solution | TestHub

For a hydrogen-like atom the energy of level n is

$E_n=-\frac{13.6,Z^2}{n^2}$ (eV)

so the photon energy for a transition $n_i\to n_f$$(n_i>n_f)$ is

$\Delta E=E_f-E_i=13.6,Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$ (eV)

Infrared radiation corresponds to a relatively small energy difference (long wavelength),

whereas ultraviolet corresponds to a large energy difference (short wavelength).

Given that $4\to 3$ is ultraviolet (i.e. a relatively large $\Delta E$ compared to small gaps),

we compare the magnitudes of $\Delta E$ for the options. For $Z=1$ (hydrogen) the numerical values are:

Option (A) $2\to 1$:

$\Delta E=13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=13.6\left(1-\frac{1}{4}\right)=13.6\times \frac{3}{4}=10.20eV$

(ultraviolet)

Option (B) $3\to 2$:

$\Delta E=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=13.6\left(\frac{1}{4}-\frac{1}{9}\right)=13.6\times \frac{9-4}{36}=13.6\times \frac{5}{36}\approx 1.8889eV$

(this lies in the visible/near-IR/near-UV region depending on exact value)

Option (C) $4\to 2$:

$\Delta E=13.6\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=13.6\left(\frac{1}{4}-\frac{1}{16}\right)=13.6\times \frac{4-1}{16}=13.6\times \frac{3}{16}=2.55eV$

(visible/near-UV region)