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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    
  1. v(34)12
     
  2. Greater than v(43)12
  3. v(43)12
     
  4. Less than v(43)12
     


Solution

The correct option is B 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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