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

Option: 1

\mathrm{4\times10^{15}Hz}
 


Option: 2

\mathrm{5\times10^{15}Hz}


Option: 3

\mathrm{1.6\times10^{15}Hz}


Option: 4

\mathrm{2.5\times10^{15}Hz}


Answers (1)

best_answer

Energy released from emission of electron

\mathrm{ E=(-3.4)-(-13.6)=10.2 \mathrm{eV} }

From photoelectric equation.

Work function,

\mathrm{ \phi=E-e V=h v }

\mathrm{ v=\frac{E-e V}{h}=\frac{(102 \times 3.57) e}{6.67 \times 10^{-34}} }

\mathrm{ v=\frac{6.63 \times 1.6 \times 10^{-19}}{6.67 \times 10^{-34}}=1.6 \times 10^{15} \mathrm{~Hz} }

Hence option 3 is correct.



 

Posted by

Shailly goel

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