In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at 10.2 V. The wavelength of light emitted by hydrogen atom when excited to the first excitation level i

Question

In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at 10.2 V. The wavelength of light emitted by hydrogen atom when excited to the first excitation level is _________ $nm$ . (Given $hc = 1245 eV nm, e = 1.6 \times 1 0^{- 19} C$ ).

Answer:

122.00

Solution:

$10.2 eV = \frac{hc}{λ} λ = \frac{1245 eV - nm}{10.2 eV} = 122.06 nm$

TestHub · Accepted Answer

​ 10.2 eV = λ hc ​ λ = 10.2 eV 1245 eV − nm ​ = 122.06 nm ​

Physics - Modern Physics Question with Solution | TestHub

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