The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV, and the stopping potential for a radiation incident on this surface 5 V. The incident radiation lies in

Option 1)

X-­ray region

Option 2)

ultra­-violet region

Option 3)

infra­-red region

Option 4)

visible region.

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 

eV_{0}=h{\nu}-{\phi}

\Rightarrow\ \;h{\nu}=eV_{0}+{\phi}=5eV+6.2eV=11.2eV

    {\lambda}=\frac{12400}{11.2}\AA=1107\AA

This lie in ultraviolet region.

Correct answer is 2.


Option 1)

X-­ray region

This is an incorrect option.

Option 2)

ultra­-violet region

This is the correct option.

Option 3)

infra­-red region

This is an incorrect option.

Option 4)

visible region.

This is an incorrect option.

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