Two identical photocathodes receive light of frequencies f_{1}\, and\, f_{2} . If the velocities of the photoelectrons (of mass m ) coming out are respectively v_{1}\, and\, v_{2} , then

Option 1)

v_{1}^{2}-v_{2}^{2}=\frac{2h}{m}(f_{1}-f_{2})

Option 2)

v_{1}+v_{2}=\left [ \frac{2h}{m}(f_{1}+f_{2}) \right ]^{\frac{1}{2}}

Option 3)

v_{1}^{2}+v_{2}^{2}=\frac{2h}{m}(f_{1}+f_{2})

Option 4)

v_{1}-v_{2}=\left [ \frac{2h}{m}(f_{1}-f_{2}) \right ]^{\frac{1}{2}}

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

 

 From Einstein's photoelectric equation 

    \frac{1}{2}mv_{1}^{2}=hf_{1}-{\phi}                            (1)

    \frac{1}{2}mv_{2}^{2}=hf_{2}-{\phi}                            (2)

Equation (1) - equation (2)

    v_{1}^{2}-v_{2}^{2}==\frac{2h}{m}\left(f_{1}-f_{2} \right )

Correct answer is 1


Option 1)

v_{1}^{2}-v_{2}^{2}=\frac{2h}{m}(f_{1}-f_{2})

This is the correct option.

Option 2)

v_{1}+v_{2}=\left [ \frac{2h}{m}(f_{1}+f_{2}) \right ]^{\frac{1}{2}}

This is an incorrect option.

Option 3)

v_{1}^{2}+v_{2}^{2}=\frac{2h}{m}(f_{1}+f_{2})

This is an incorrect option.

Option 4)

v_{1}-v_{2}=\left [ \frac{2h}{m}(f_{1}-f_{2}) \right ]^{\frac{1}{2}}

This is an incorrect option.

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