Dual Nature of Matter

JEE Main level, single correct, medium

177 of 200

Question 177 of 200

The magnetic field associated with a light wave is given, at the origin, by B=B0[sin⁡⁡(3.14×107)ct+sin⁡⁡(6.28×107)ct].B={B}_{0}\left[\sin ⁡\left(3.14\times {10}^{7}\right)ct+\sin ⁡\left(6.28\times {10}^{7}\right)ct\right]. If this light falls on a silver plate having a work function of 4.7eV,4.7eV, what will be the maximum kinetic energy of the photoelectrons?

(c=3×108ms−1,h=6.6×10−34Js)(c=3\times {10}^{8} m {s}^{-1},h=6.6\times {10}^{-34} J s)

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Answer: (B) 7.72eV7.72eV

B=B0[sin⁡⁡(3.14×107ct)+sin⁡⁡(6.28×107ct)]B={B}_{0}\left[\sin ⁡\left(3.14\times {10}^{7}ct\right)+\sin ⁡\left(6.28\times {10}^{7}ct\right)\right]

Maximum KEKE will associate with the maximum frequency

fmax⁡⁡=6.28×107c2π=107×3×108Hz=3×1015Hz{f}_{\max ⁡}=\dfrac{6.28\times {10}^{7}c}{2\pi }={10}^{7}\times 3\times {10}^{8}Hz =3\times {10}^{15}Hz

KEmax=hfmax−ϕK{E}_{max}=h{f}_{max}-\phi

=[6.63×10−34×3×10151.6×10−19−4.7eV]=\left[\dfrac{6.63\times {10}^{-34}\times 3\times {10}^{15}}{1.6\times {10}^{-19}}-4.7eV\right]

=7.72eV=7.72eV

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