JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 11)
The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is :
194.8 nm
490.7 nm
913.3 nm
121.8 nm
Explanation
$$\Delta$$E = 10.2 eV
$${{hc} \over \lambda }$$ = 10.2 eV
$$\lambda$$ = $${{hc} \over {(10.2)e}}$$
= $${{12400} \over {10.2}}\mathop A\limits^o $$
= 121.56 nm
$$ \simeq $$ 121.8 nm
$${{hc} \over \lambda }$$ = 10.2 eV
$$\lambda$$ = $${{hc} \over {(10.2)e}}$$
= $${{12400} \over {10.2}}\mathop A\limits^o $$
= 121.56 nm
$$ \simeq $$ 121.8 nm
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