Question 2.13
NCERT Class XI Chemistry
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?
Video Explanation
(detailed solution given after this video)
Please watch this youtube video for a quick explanation of the solution:
Solution in Detail
Given $\displaystyle n_i = 4\text{, } n_f = 2$
visible Balmer emission because $\displaystyle n_f = 2$
Calculate wave number first:
$\displaystyle \overline{\nu} = 1.09677 \times 10^7 \bigg(\frac{1}{n_i^2} - \frac{1}{n_f^2}\bigg)\text{(/m)}$
$\displaystyle = 1.09677 \times 10^7 \bigg(\frac{1}{4^2} - \frac{1}{2^2}\bigg) \text{(/m)}$
$\displaystyle = -2.06 \times 10^6 \text{ /m}$
-ve sign means emission. can be ignored.
but wavelength $\displaystyle \lambda = \frac{1}{\overline{\nu} }$
$\displaystyle = \frac{1}{2.06 \times 10^6 } \text{ m}$
$\displaystyle = 486 \times 10^{-9} \text{ m} = 486 \text{ nm }\underline{Ans}$
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