Q. The wavelength of line emitted when an electron jumps from an outer orbit to inner orbit which possess an energy difference of $3eV$ will be:

NTA AbhyasNTA Abhyas 2020 Report Error

A

$3660Å$

B

$3620Å$

C

$4140 \, Å$

D

$4560Å$

Solution:

Use $\textit{E}=\textit{hc}/\textit{\lambda }$
Or $\textit{E}\left(\text{ev}\right)=\frac{1 2 4 2 0}{\textit{\lambda } \left(\text{\mathring{A} }\right)}$
$\Rightarrow \textit{\lambda }\left(\mathring{A} \right)=\frac{1 2 4 2 0}{3}=4140\text{\mathring{A} }$

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