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When photons of energy 4.25 eV strike th...

When photons of energy `4.25 eV` strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy `T_(A)` eV and De-broglie wavelength `lambda_(A)`. The maximum energy of photoelectron liberated from another metal B by photon of energy 4.70 eV is `T_(B) = (T_(A) - 1.50) eV` if the de Brogle wavelength of these photoelectrons is `lambda_(B) = 2 lambda_(A) `, then

A

The work function of A is 2.25 eV

B

The work function of B is `4.20 eV`

C

`T_(A) = 2.00 eV`

D

`T_(B) = 2.75 eV`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
(1) `lambda_(A) = (h)/(sqrt(2mT_(A)))` , (2) `lambda_(B) = (h)/(sqrt(2mT_(B)))`
(3) `T_(B) = T_(A) - 1.5 eV` , (4) `lambda_(B) = 2lambda_(A)`
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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 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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