Question

# When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and deBroglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB=(TA−1.5) eV. If the deBroglie wavelength of these photoelectrons is λB=2λA, then:

A
Work function of A is 2.25eV
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B
Work function of B is 4.8eV
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C
TA=3 eV
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D
TB=2.75 eV
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Solution

## The correct option is C Work function of A is 2.25eVFor metal A,de-broglie wavelength (λA)=hmvA−−−−−−1⇒VA=hmλA4.25−WA=12×m×h2m2×A2⇒4.25−WA=h22mλA2−−−−−−2For metal B,4.7−WB=h22mλA2−15−−−−−3and λB=hmvB⇒2λA=hmvB⇒VB=h2mλA−−−−−4So,from difference in kinetic energy,⇒−TB+TA=1.5−−−−xand solving (i),(ii),(iii),(iv) and (v) simultaneously we find, WA=2.25eV.

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