JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 24)
An excited He+
ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a
transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of
wavelength $$\lambda $$, energy $$E = {{1240\,eV} \over {\lambda (in\,nm)}}$$) :
n = 4
n = 7
n = 5
n = 6
Explanation
$$\Delta {E_n} = - {{{E_0}{Z^2}} \over {{n^2}}}$$
Let it start from n to m and from m to ground.
Then $$13.6 \times 4\left| {1 - {1 \over {{m^2}}}} \right| = {{hc} \over {30.4\,nm}}$$
$$ \Rightarrow 1 - {1 \over {{m^2}}} = 0.7498 \Rightarrow 0.25 = {1 \over {{m^2}}}$$
$$ \therefore $$ m = 2, and now $$13.6 \times 4\left( {{1 \over 4} - {1 \over {{n^2}}}} \right) = {{hc} \over {108.5 \times {{10}^{ - 9}}}}$$
n = 5
Let it start from n to m and from m to ground.
Then $$13.6 \times 4\left| {1 - {1 \over {{m^2}}}} \right| = {{hc} \over {30.4\,nm}}$$
$$ \Rightarrow 1 - {1 \over {{m^2}}} = 0.7498 \Rightarrow 0.25 = {1 \over {{m^2}}}$$
$$ \therefore $$ m = 2, and now $$13.6 \times 4\left( {{1 \over 4} - {1 \over {{n^2}}}} \right) = {{hc} \over {108.5 \times {{10}^{ - 9}}}}$$
n = 5
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