The threshold frequency of a metal is f0 . When the light of frequency 2f0 is incident on the metal plate, the maximum velocity of photoelectrons is v1 . When the frequency of incident radiation is increased to 5f0 , the maximum velocity of photoelectrons emitted is v2 . The ratio of v1 to v2 is:

Eph=W+12mv2⇒hv=hv0+12mv2⇒h2f0=hf0+12mv12 (1)&h5f0=hf0+12mv2 2 (2)⇒hf04f0=12mv1212mv22⇒14=v1v22⇒v1v2=12

The threshold frequency of a metal is f0 . When the light of frequency 2f0 is incident on the metal plate, the maximum velocity of photoelectrons is v1 . When the frequency of incident radiation is increased to 5f0 , the maximum velocity of photoelectrons emitted is v2 . The ratio of v1 to v2 is:

Eph=W+12mv2⇒hv=hv0+12mv2⇒h2f0=hf0+12mv12 (1)&h5f0=hf0+12mv2 2 (2)⇒hf04f0=12mv1212mv22⇒14=v1v22⇒v1v2=12

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The threshold frequency of a metal is f₀ . When the light of frequency 2f₀ is incident on the metal plate, the maximum velocity of photoelectrons is v₁ . When the frequency of incident radiation is increased to 5f₀ , the maximum velocity of photoelectrons emitted is v₂ . The ratio of v₁ to v₂ is:

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$\frac{v_{1}}{v_{2}} = \frac{1}{4}$

$\frac{v_{1}}{v_{2}} = \frac{1}{16}$

$\frac{v_{1}}{v_{2}} = \frac{1}{2}$

$\frac{v_{1}}{v_{2}} = \frac{1}{8}$

answer is D.

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Detailed Solution

$E_{ph} = W + \frac{1}{2} {mv}^{2} \Rightarrow hv = {hv}_{0} + \frac{1}{2} {mv}^{2}$
$\Rightarrow h (2 f_{0}) = {hf}_{0} + \frac{1}{2} {mv}_{1}^{2} (1)$
$& h (5 f_{0}) = {hf}_{0} + \frac{1}{2} {mv}_{2}^{2} (2)$
$\Rightarrow \frac{{hf}_{0}}{4 f_{0}} = \frac{\frac{1}{2} {mv}_{1}^{2}}{\frac{1}{2} {mv}_{2}^{2}} \Rightarrow \frac{1}{4} = {(\frac{v_{1}}{v_{2}})}^{2} \Rightarrow \frac{v_{1}}{v_{2}} = \frac{1}{2}$