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Chemistry Question on Dual nature of matter

The velocity of the most energetic electrons emitted from a metallic surface is doubled when the frequency v of the incident radiation is doubled. The work function of this metal is

A

hv4\frac{ h v }{ 4 }

B

hv3\frac{ h v }{ 3 }

C

hv2\frac{ h v }{ 2 }

D

2hv3\frac{ 2h v }{ 3 }

Answer

2hv3\frac{ 2h v }{ 3 }

Explanation

Solution

12mv212m(2v)2=hv−ϕh(2v)−ϕ\frac{ \frac{1}{2} mv^2 }{ \frac{1}{2} m (2 v)^2 } = \frac{ h v - \phi }{ h ( 2 v ) - \phi } or 14=hv−ϕ2hv−ϕ\frac{1}{ 4 } = \frac{ h v - \phi }{ 2 hv - \phi } or ϕ=2hv3\phi = \frac{ 2 hv }{ 3 }