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Radiation of frequency \( v \) is incident on a photosensitive metal. The maximum kinetic energy of photoelectric is \( E \). When the frequency of the incident radiation is doubled, what is the maximum kinetic energy of the photoelectrons?(a) \( 4 E \)(b) \( 2 E \)(c) \( E+h v \)(d) \( E-h v \) |
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Answer» Answer: (c) \(E+h\nu\) Energy of incident ray = threshold energy + (K.E)max \(h\nu=(h\nu_0)+E....(1)\) \(\Rightarrow E=h\nu-h\nu_0\) In the second case, threshold energy remains the same \(h(2\nu)=(h\nu_0)+(K.E)_{max}....(2)\) By (1) & (2) eqn (K.E)max=\([h\nu-h\nu_0]+h\nu\) \(K.E = E+h\nu\) |
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