What is the wavelength of light emitted when the electron in a hydrogen atom undergoes the transition from an energy level with n = 4 to an energy level with n = 2?

Asked by Pragya Singh | 1 year ago |  99

##### Solution :-

The $$n_{i} = 4\; to\; n_{f}= 2$$ transition results in a spectral line of the Balmer series. The energy involved in this transition can be calculated using the following expression:

$$=2.18\times 10^{-18}[\dfrac{1}{n_{i}^{2}}-\dfrac{1}{n_{f}^{2}}]$$

Substituting these values in the expression for E:

$$E=2.18\times 10^{-18}[\dfrac{1}{4^{2}}-\dfrac{1}{2^{2}}]\\ \\ =2.18\times 10^{-18}[\dfrac{1-4}{16}]\\ \\ =2.18\times 10^{-18}\times (-\dfrac{3}{16})\\ \\ E=-(4.0875\times 10^{-19}J)$$

Here, the -ve sign denotes the emitted energy.

Wavelength of the emitted light $$(\lambda )=\dfrac{hc}{E}$$

(Since $$E=\dfrac{hc}{\lambda })$$

Substituting these values in the expression for $${\lambda }$$

$$\lambda =\dfrac{(6.626\times 10^{-34})(3\times 10^{8})}{(4.0875\times 10^{-19})}$$

$$=4.8631\times 10^{-7}\, m\\ \\$$

$$\lambda =486.31\times 10^{-9}\, m\\ \\$$

$$=486\, nm$$

Answered by Pragya Singh | 1 year ago

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