The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface.  If the incident frequency is increased to 3n, the maximum velocity of the ejected photoelectrons will be :

 

Option 1)

 less than \sqrt{3} V

Option 2)

v

Option 3)

 more than \sqrt{3} V

Option 4)

equal to \sqrt{3} V

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

 

 

K.E_{max}=h{\nu }-{\phi }

\frac{1}{2}mv^{2}=hn-{\phi}                    ......(1)

\frac{1}{2}mv'^{2}=3hn-{\phi}=3(nh-{\phi})+2{\phi}

    =3.\frac{1}{2}mv^{2}+2{\phi}                (From equation (1))

v'^{2}=3v^{2}+\frac{2{\phi}}{m}

v'^{2}=\sqrt{3v^{2}+\frac{2{\phi}}{m}}>\sqrt{3}v


Option 1)

 less than \sqrt{3} V

This is an incorrect option.

Option 2)

v

This is an incorrect option.

Option 3)

 more than \sqrt{3} V

This is the correct option.

Option 4)

equal to \sqrt{3} V

This is an incorrect option.

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