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When photon of energy 25eV strike the ...

When photon of energy `25eV` strike the surface of a metal A, the ejected photelectron have the maximum kinetic energy photoelectrons have the maximum kinetic energy `T_(A)eV` and de Brogle wavelength `lambda_(A)` .The another kinetic energy of photoelectrons liberated from another metal B by photons of energy `4.76 eV` is `T_(B) = (T_(A) = 1.50) eV` .If the de broglie wavelength of these photoelectrons is `lambda_(B) = 2 lambda_(A)` then
i.` (W_(B))_(A) = 2.25 eV II. (W_(0))_(B) = 4.2 eV`
III`T_(A) = 2.0 eV IV. T_(B) = 3.5 eV`

A

I,II

B

II,III,IV

C

I,II,III

D

I,II,III,IV

Text Solution

Verified by Experts


`W_(A) + T_(A) = 4.25 rArr W_(A) = 2.25 eV`
`W_(A) + T_(B) = 4.7 rArr W_(B) = 4.2 eV`
`lambda_(B) = 2 lambda_(A)`
`rArr KE_(A) = 4KE_(B)`
`rArr T_(A) = 4(T_(B))`
`rArr T_(A) = 2.0 eV`
`T_(A) = 0.5 eV`
Hence option c
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When photons of energy 4.25eV strike the surface of a metal A , the ejected photoelectrons have maximum kinetic energy T_(A) (expressed in eV ) and de-Broglie wavelength lambda . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70eV is T_(B)=T_A-1.50 eV . If the de-Broglie wavelength (in eV ) of these photoelectrons is lambda_(B)=2 lambda_(A) then find T_(B) (in eV).

When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy lamda_(A). The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is T_(B)=(T_(A) -1.50) eV. If the de Broglie wavelength of these photoelectrons is lamda_(B)=2lamda_(A), then select the correct statement statement (s)

Knowledge Check

  • When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, T_(A) (expressed in eV) and de Broglie wavelength lambda_(A) . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20 eV is T_(B)=T_(A)-1.50 eV . If the de Broglie wavelength of these photoelectrons is lambda_(B)=2lambda_(A) , then which is not correct?

    A
    The work function of A is 2.25 eV
    B
    The work function of B is 3.70 eV
    C
    `T_(A)=2.00 eV`
    D
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  • When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, T_A (expressed in eV) and deBroglie wavelength lambda_A . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20V is T_B = T_A -1.50eV . If the deBroglie wavelength of those photoelectrons is lambda_B = 2lambda_A then

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    the work function of A is 2.25 eV
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    A
    The work function of A is 1.50eV
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    The work function of B is 4.0 eV
    C
    `T_(A)=3.2eV`
    D
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