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Calculate the energy and frequency of the radiation emitted when an electron jumps from n= 3 to n= 2 in a hydrogen atom.

Answers (1)

When we use the Rydberg's Equation for hydrogen as well as an hydrogen like atom:

$\frac{1}{\lambda }=R_{H}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{f}^{2}} \right )$

Where,

$\lambda$ Represents the wavelength of radiation

$R_{H}$ Is the Rydberg's Constant i.e. $1.097\times 10^{7}m^{-1}$

$n_{f}$ = Higher energy level = 3

$n_{i}$ = Lower energy level = 2

$\frac{1}{\lambda } = 1.097\times 10^7 m^{-1}\times \left ( 1/2^2 -1/3^2\right )$

$\frac{1}{\lambda } = 1.097\times 10^7 m^{-1}\times \frac{5}{36}$

$\lambda = 6.7\times 10^{-7} m$

$\lambda = \frac{c}{\nu }$

where,

$\lambda$ represents the wavelength of the light i.e. $6.7\times 10^{-7}m$

c represents the speed of light i.e. $3\times 10^{8}m/s$

$\nu$ represents the frequency of light i.e. $\frac{c}{\lambda } = \frac{3\times 10^8 m/s}{6.7\times 10^{-7}m} = 0.4\times 10^{15} s^{-1}$

$E = \frac{6.6\times 10^{-34} \times 3\times 10^8}{6.7\times 10^{-7}m}$

Hence, the energy of radiation emitted is 3.028 x 10^-19J and the frequency is 0.4 x 10¹⁵ s^-1

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