JEE MAIN - Physics (2022 - 27th June Morning Shift - No. 17)

A hydrogen atom in its ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of :

(Given, Planck's constant = 6.6 $$\times$$ 10^$$-$$34 Js).

2.10 $$\times$$ 10^$$-$$34 Js

1.05 $$\times$$ 10^$$-$$34 Js

3.15 $$\times$$ 10^$$-$$34 Js

4.2 $$\times$$ 10^$$-$$34 Js

$$ - 13.6 + 10.2 = {{ - 13.6} \over {{n^2}}}$$

$$ \Rightarrow {{13.6} \over {{n^2}}} = 3.4$$

$$ \Rightarrow n = 2$$

$$ \Rightarrow \Delta L = 2 \times {h \over {2\lambda }} - 1 \times {h \over {2\lambda }}$$

$$ = {h \over {2\lambda }}$$

$$ \Rightarrow \Delta L \simeq 1.05 \times {10^{ - 34}}$$ Js