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Question

In a photoemissive cell with executing 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

A
v(34)12
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B
v(43)12
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C
Less than v(43)12
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D
Greater than v(43)12
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Solution

The correct option is D Greater than v(43)12
hvW0=12mv2maxhcλhcλ0=12mv2max
hc(λ0λλλ0)=12mv2maxvmax=2hcm(λ0λλλ0)
When wavelength is λ and velocity is v, then
v=2hcm(λ0λλλ0) …. (i)
When wavelength is 3λ4 and velocity is v' then
v= 2hcm[λ0(3λ4)(3λ4)×λ0] ….(ii)
Divide equation (ii) by (i), we get
vv= [λ0(3λ4)]34λλ0×λλ0λ0λ
v=v(43)12[λ0(3λ4)]λ0λi.e.v>v(43)12


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