What is the work function of the metal if the light of wavelength 400\AA generates photoelectrons of velocity 6\times 10^{5}ms^{-1} from it ?

(Mass of electron = 9\times 10^{-31}kg

Velocity of light = 3\times 10^{8}ms^{-1}

Planck's constant = 6.626\times 10^{-34}Js

Charge of electron = 1.6\times 10^{-19}JeV^{-1})

  • Option 1)

    0.9eV

  • Option 2)

    3.1eV

  • Option 3)

    2.1eV

  • Option 4)

    4.0eV

Answers (1)
A admin

 

De-broglie wavelength -

\lambda = \frac{h}{mv}= \frac{h}{p}

- wherein

where m is the mass of the particle

v its velocity 

p its momentum

 

 

Total energy of elctron in nth orbit -

E_{n}= -13.6\: \frac{z^{2}}{n^{2}}eV

Where z is atomic number

As we have learned in work function

h\nu = \phi +h\nu ^{0}

\frac{1}{2}mv^{2}= hc\left [ \frac{1}{\lambda }-\frac{1}{\lambda _{0}} \right ]

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

\phi = \frac{6.626\times 10^{-34}\times 3\times 10}{4000\times 10^{-10}}--\frac{1}{2}\times 9\times 10^{-31}\times \left ( 6\times 10^{5} \right )^{2}

\phi = 3.35\times 10^{-19}J

\phi = 2 .1eV

 

 


Option 1)

0.9eV

Option 2)

3.1eV

Option 3)

2.1eV

Option 4)

4.0eV

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