The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state l, would be (Rydberg constant = 1.097×107m−1)

# The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state l, would be (Rydberg constant = $1.097×{10}^{7}{m}^{-1}$)

1. A

91 nm

2. B

192 nm

3. C

406 nm

4. D

$9.2×{10}^{-8}nm$

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### Solution:

We have    $\Delta \stackrel{~}{E}={R}_{\infty }\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)={R}_{\infty }\left(\frac{1}{{l}^{2}}-\frac{1}{{\infty }^{2}}\right)={R}_{\infty };\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\lambda =\frac{1}{\stackrel{~}{{R}_{\infty }}}=0.91×{10}^{-7}m=91\text{\hspace{0.17em}\hspace{0.17em}}nm$

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