Question 274: Atoms & Nuclei

Question

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 $$E = {{1240\,eV} \over {\lambda (in\,nm)}}$$) :

Accepted Answer

$$\Delta {E_n} =  - {{{E_0}{Z^2}} \over {{n^2}}}$$
Let it start from n to m and from m to ground.
Then $$13.6 	imes 4\left| {1 - {1 \over {{m^2}}}} ight| = {{hc} \over {30.4\,nm}}$$
$$ \Rightarrow 1 - {1 \over {{m^2}}} = 0.7498 \Rightarrow 0.25 = {1 \over {{m^2}}}$$
$$ 	herefore $$ m = 2, and now $$13.6 	imes 4\left( {{1 \over 4} - {1 \over {{n^2}}}} ight) = {{hc} \over {108.5 	imes {{10}^{ - 9}}}}$$
n = 5

Answer

n = 4

Answer

n = 7

Answer

n = 5

Answer

n = 6

Atoms & Nuclei

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 $\lambda , energy E = {{1240\,eV} \over {\lambda (in\,nm)}}$) :

Choose the correct answer: