Question

A difference of $2.3eV$ separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Answer 1

Given: The difference in energy between the two energy levels is $2.3eV$.

To find: The frequency of radiation emitted when the atom makes a transition between the energy levels.

The energy between the energy levels is

$E=2.3eV\Rightarrow 2.3\times 1.6\times {10}^{-19}J$

The frequency of the radiations emitted during transition is:

$E=h\nu$

$\nu =\frac{h}{E}$

$\Rightarrow \frac{6.626\times {10}^{-34}}{2.3\times 1.6\times {10}^{-19}}$

$\Rightarrow 5.5\times {10}^{14}Hz$