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  • Confused! kindly explain, The photoelectric threshold wavelength of silver is 3250 x 10-10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 x 10-10 m is :(Given h = 4.14 x 10-15 eVs and c = 3 x 108

The photoelectric threshold wavelength of silver is 3250 x 10-10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 x 10-10 m is :

(Given h = 4.14 x 10-15 eVs and c = 3 x 108 ms-1)

  • Option 1)

    \approx 6 \times 10^5 \text{ms}^{ - 1}

  • Option 2)

    \approx 0.6 \times 10^6 \text{ms}^{ - 1}

  • Option 3)

    \approx 61 \times 10^3 \text{ms}^{ - 1}

  • Option 4)

    \approx 0.3 \times 10^6 \text{ms}^{ - 1}

 

Answers (1)

 

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

 

 maximum kinetic energy =\frac{hc}{\lambda }=\o

K.E=\frac{hc}{\lambda }=\frac{hc}{\lambda_{o} }=hc(\frac{1}{\lambda }-\frac{1}{\lambda _{o}})=\frac{1}{2}mv^{2}

V=\sqrt{\frac{2hc}{m}.\left ( \frac{1}{\lambda }-\frac{1}{\lambda _{o}} \right )}

\lambda =6.67\times 10^{-34}J-s\\ c=3\times 10^{18}m/s\\ \lambda =2536\times 10^{-10}m\\ m=9.1\times 10^{-31}kg

Putting all these values we get 

V=0.3\times 10^{6}m/s

 


Option 1)

\approx 6 \times 10^5 \text{ms}^{ - 1}

This solution is incorrect 

Option 2)

\approx 0.6 \times 10^6 \text{ms}^{ - 1}

This solution is incorrect 

Option 3)

\approx 61 \times 10^3 \text{ms}^{ - 1}

This solution is incorrect 

Option 4)

\approx 0.3 \times 10^6 \text{ms}^{ - 1}

This solution is correct 

Posted by

Sabhrant Ambastha

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