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When photons of energy 4.25 eV strike th...

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)

Text Solution

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The correct Answer is:
`A=2.25eV,B=4.20 eV, T_(A)=2.00eV.`
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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 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

Knowledge Check

  • When photon of energy 4.0eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy T_(A) eV and de-Broglie wavelength lamda_(A) . The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50eV is T_(B)= (T_(A)-1.50)eV . If the de-Broglie wavelength of these photoelectrons lamda_(B)=2lamda_(A) , then choose the correct statement(s)

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

  • 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
    `T_(B)=2.75 eV`

  • 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

    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
    `T_(B)=2.75 eV`

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