Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited state to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the materials is:

  • Option 1)

    4\times 10^{15}\: \: Hz

  • Option 2)

    5\times 10^{15}\: \: Hz

  • Option 3)

    1.6\times 10^{15}\: \: Hz

  • Option 4)

    2.5 \times 10^{15}\: \: Hz

 

Answers (1)

As we learnt in

Conservation of energy -

h\nu = \phi _{0}+\frac{1}{2}mv^{2}_{max}

h\nu = h\nu _{0}+\frac{1}{2}mv^{2}_{max}

 

h\left ( \nu-\nu _{0} \right )=\frac{1}{2}mv^{2}_{max}

where, h - Plank's constant\ \nu - Frequency\ \nu_{0} - threshold\ frequency\ \phi_{0}- work function

- wherein

 

 

 

Amount of energy radiated when electron jump from n = 2 to n = 1

\Delta E=13.6\:(1-\frac{1}{4})\:=\:10.2\:eV

Then  eV_{s}\:=\:\frac{hc}{\lambda }-\phi

\phi = 10.2\:eV-3.57\:eV

= 6.63\:eV\:=\:h\upsilon

Threshold frequency = \frac{6.63\times1.6\times10^{-19}}{6.62\times10^{-34}}=\:1.6\times10^{15}\:HZ


Option 1)

4\times 10^{15}\: \: Hz

Incorrect

Option 2)

5\times 10^{15}\: \: Hz

Incorrect

Option 3)

1.6\times 10^{15}\: \: Hz

Correct

Option 4)

2.5 \times 10^{15}\: \: Hz

Incorrect

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