

InterviewSolution
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For photoelectric emission from certain metal the cutoff frequency is v. If radiation of frequency 2 v impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass):1. \(\sqrt {hv/(2m)}\)2. \(\sqrt {hv/m}\)3. \(\sqrt {2hv/m}\)4. \(2\sqrt {hv/m}\) |
Answer» Correct Answer - Option 4 : \(2\sqrt {hv/m}\) CONCEPT:
K = hν -ϕ ν is the frequency of light striking metal surface, h is Planck's constant,ϕ is work function.
ϕ = hν0 whereν0 is the cutoff frequencyfor a light ray to emit electron out of given metal. Kinetic Energy: The energy possed by a body in motion is known as Kinetic Energy. \(K = \frac{1}{2} mv^2 \) m is mass of the body, v is the velocity of the body CALCULATION: Given Cuttoff frequency = ν Frequency of light ray striking = 2ν The energy of each photon falling = 2hν the Maximum kinetic energy K obtained from the metal havingcutoff frequency is v. K = 2hν - hν (By Einestein Equation) ⇒ K = hν ⇒ \(K = \frac{1}{2} mv^2 = h\nu\) ⇒\(v^2 = \frac{ 2h\nu}{m}\) ⇒\(v =\sqrt{\frac{ 2h\nu}{m}} \) |
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