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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