Question

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

Answer 1

Explanation:

..........,

Answer 2

The energy and frequency of the radiation emitted when an electron jumps from n=3 to n=2 in hydrogen atom is $2.9\times 10^{-19}J$ and $0.4\times 10^{15}s^{-1}$

Explanation:

Using Rydberg's Equation for hydrogen and hydrogen like atom:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant  =$1.097\times 10^7 m^{-1}$

$n_f$ = Higher energy level  = 3

$n_i$ = Lower energy level = 2

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

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

$\frac{1}{\lambda}=0.15\times 10^7 m^{-1}$

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

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

where,

$\lambda$ = wavelength of the light =$6.7\times 10^{-7} m$

c = speed of light = $3\times 10^8m/s$

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

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

$E=2.9\times 10^{-19}J$

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