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Question

In a photoemissive cell with exciting wavelength λ , the fastest electron has speed v. If the exciting wavelength is changed to 3λ4 , the speed of the fastest emitted electron will be


Your Answer
A
v(34)1/2
Your Answer
B
v(43)1/2
Your Answer
C
Less than v(43)1/2
Correct Answer
D
Greater then v(43)1/2

Solution

The correct option is B Greater then v(43)1/2
From E=W0+12mv2maxvmax=2Em2W0m (where E=hcλ)
If wavelength of incident light charges from λ to 3λ4 (decreases)
Let energy of incident light charges from E to E’ and speed of fastest electron changes from v to v’ then
v=2Em2W0m  . . . . . . .(i)  and v=2Em2W0m   . . . . . .(2)
As E1λE=43E hence v=2(43E)m2W0mv=(43)122Em2W0m(43)12
v=(43)12 2Em2W0m(43)12  > v

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