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Q.

When photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is TB=TA−1.5eV. If the de-Broglie wavelength of these photoelectrons λB=2λA, then the work function of metal B is:

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a

1.5 eV

b

4eV

c

2eV

d

3eV

answer is B.

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Detailed Solution

Relation between De – Broglie wavelength and K.E. isλ=h2KEme⇒λ∝1KE ⇒ λAλB=KEBKEA⇒12=TA−1.5TA⇒TA=2eV ∴KEB=2−1.5=0.5eVNow, for metal B work function is ϕB = EB−KEB ⇒ ϕB=4.5−0.5=4eV

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When photon of energy 4.0 eV strikes 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 photon of energy 4.50 eV is TB=TA−1.5eV. If the de-Broglie wavelength of these photoelectrons λB=2λA, then the work function of metal B is: