When photons of energy 4.25 {eV} strike the surface of a metal {A}, the ejected photoelectrons have maximum kinetic energy {T}_{ {A}} {eV} and de-Broglie wavelength \lambda_{ {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 \lambda_{B}=2 \lambda_{A} then

the work function of A is 2.25 {eV}

the work function of {B} is 4.20 {eV}

JEE Advanced 2020 Full Test 4 Paper 1

Question : 7 of 54

Marks: +1, -0

When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TAeV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70eV is TB=(TA−1.50)eV. If the de-Broglie wavelength of these photoelectrons is λB=2λA then

the work function of A is 2.25eV
the work function of B is 4.20eV
TA=2.00eV
TB=2.75eV

Validate

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