Manipal Entrance Test 2014 Physics Paper

From E=W0+

1

2

mvmax2⇒vmax=√

2E

m

−

2W0

m

(where E=

hc

λ

)
If wavelength of incident light charges from λ to

3λ

4

(decreases).
Let energy of incident light charges from E to E′ and speed of fastest electron changes from v to v, then
v=√

2E

m

−

2W0

m

.....(i)
and v=√

2E′

m

−

2W0

m

....(ii)
As E∝

1

λ

⇒E′=

4

3

E
Hence, v=√

2( 4 3 E)

m

−

2W0

m

⇒ v′=(

4

3

)1∕2√

2E

m

−

2W0

m( 4 3 ) 1 2

⇒ v′=(

4

3

)1∕2 X=√

2E

m

−

2W0

m( 4 3 ) 1 2

>v
So v′>(

4

3

)1∕2v