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Question

A small piece of Cesium metal having work function (ϕ=1.9 eV) is kept at a distance of 20 cm from a large metal plate having a charge density of 1.0×109 C m2 on the surface facing the small Cesium piece. A monochromatic light of wavelength 400 nm is incident on the Cesium piece. Find the minimum and the maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present.(hc=1240 eV nm)

A
22.6 eV ; 23.8 eV
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B
0 eV ; 11.3 eV
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C
32 eV ; 48 eV
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D
9 eV ; 10 eV
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Solution

The correct option is A 22.6 eV ; 23.8 eV
Given,

σ=1×109 C m2 ; ϕCs=1.9 eVd=20 cm ; λ=400 nm

The electric potential difference between the charged plates is,

V=E×d=σε0×d [ E=σε0]

=1×109×20×1028.85×101222.6 V

From, Einstein's photoelectric equation,

eV0=Eϕ0

eV0=hcλϕ0=12404001.9

V0=3.11.9=1.2 V

As V0 is much less than V

Hence, the minimum energy required to reach the charged plate will be,

KEmin=eV0=22.6 eV

KE of the emittted electron from the ceisium plate lies from zero to KEmax but the pd between the plates will provide the extra energy.

Hence, KEmax=eV0+eV=1.2+22.6=23.8 eV

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (A) is the correct answer.

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