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PhysicsInteger+4Question 38 of 60

IIT-JEE 2011 · Paper 2, Physics question 38

A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-space. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is A×10ZA\times10^{Z} (where 1<A<101<A<10). The value of 'ZZ' is

Answer and worked solution

Official answer: 7

Photon energy 1240200=6.2\dfrac{1240}{200}=6.2 eV, so emission stops when the sphere's potential reaches 6.2−4.7=1.56.2-4.7=1.5 V.

Charge. Q=4πε0rV=0.01×1.59×109=1.67×10−12Q=4\pi\varepsilon_0rV=\dfrac{0.01\times1.5}{9\times10^9}=1.67\times10^{-12} C.

Count. n=Qe=1.67×10−121.6×10−19≈1.04×107n=\dfrac Qe=\dfrac{1.67\times10^{-12}}{1.6\times10^{-19}}\approx1.04\times10^7, so Z=7Z=7.

Answer: 7

Topics:Photoelectric effectElectric potential
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