The work function of metal is 1 eV. Light of wavelength 3000 Å is incident on this metal surface. The velocity of emitted photo-electrons will be
A
10 m/sec
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B
1×103m/sec
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C
1×104m/sec
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D
1×106m/sec
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Solution
The correct option is D1×106m/sec E=W0+Kmax;E=123753000=4.125eV ⇒Kmax=E−W0=4.125eV−1eV=3.125eV ⇒12mv2max=3.125×1.6×10−19J ⇒vmax=√2×3.125×1.6×10−199.1×10−31=1×106m/s