Question

Calculate the energy of the photon emitted when an excited electron of a hydrogen atom relaxes from n = 7 to n = 1

Answer 1

The energy of the photon will be  -2.10 × $10^{-18}$.

The energy that a single photon carries is known as photon energy. Energy is inversely correlated with wavelength because it is directly proportional to the electromagnetic frequency of something like the photon. The energy of a photon increases with its frequency.

The energy of the photon can be determined by the formula:

ΔE = R (1/$n^{2} _{2}$ - 1/ $n^{2} _{1}$ )

where, E is energy and n is the level of the energy level.

Given data:

$n_{1}$ = 1

$n_{2}$ = 7

put the given data in above equation.

ΔE = R (1/$14$ - 1/ $1$ )

ΔE= 2.17 × $10^{-18}$ (-0.97)

ΔE = -2.10 × $10^{-18}$.

Therefore, the energy of the photon will be  -2.10 × $10^{-18}$.

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