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 \underline\hspace2cm . [Use : Bohr radius, a=52.9 pm Rydberg constant, R_H=2.2 \times 10^-18 ~J Planck's constant, h=6.6 \times 10^-34 ~J ~s Speed of light, c=3 \times 10^8 ~m ~s^-1 ]

Structure of Atom PYQ — JEE Advanced 2023 · Shift 2 · Q11 with Solution

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Correct answer: 30

For a hydrogen-like ion, $r_n = \frac{n^2 a_0}{Z}$ where for $\mathrm{He}^+$ , $Z=2$ and $a_0=52.9\,\mathrm{pm}$ .

So, $r_n = \frac{n^2(52.9)}{2} = n^2(26.45)\,\mathrm{pm}$

Given radii:

Now, $105.8 = 4 \times 26.45 \implies n_i^2 = 4 \implies n_i=2$

and $26.45 = 1 \times 26.45 \implies n_f^2 = 1 \implies n_f=1$

Thus, the transition is: $n=2 \to n=1$

For a hydrogen-like ion, $E_n = -\frac{Z^2 R_H}{n^2}$ Here $Z=2$ , so $E_n = -\frac{4R_H}{n^2}$

Hence, $E_1 = -4R_H$ $E_2 = -\frac{4R_H}{4} = -R_H$

Energy emitted: $\Delta E = E_2 - E_1 = (-R_H)-(-4R_H)=3R_H$

Using $R_H = 2.2\times 10^{-18}\,\mathrm{J}$ , $\Delta E = 3(2.2\times 10^{-18}) = 6.6\times 10^{-18}\,\mathrm{J}$

$\Delta E = \frac{hc}{\lambda}$ So, $\lambda = \frac{hc}{\Delta E}$

Substitute values: $\lambda = \frac{(6.6\times 10^{-34})(3\times 10^8)}{6.6\times 10^{-18}}$

$\lambda = 3\times 10^{-8}\,\mathrm{m}$

Convert to nm: $3\times 10^{-8}\,\mathrm{m} = 30\times 10^{-9}\,\mathrm{m} = 30\,\mathrm{nm}$

The wavelength of the emitted photon is: $\boxed{30}$ (in nm)