Question:

Radiation of frequency \(2ν_0\) is incident on a metal with threshold frequency \(ν_0\). The correct statement of the following is ______.

Updated On: Oct 30, 2024
  • No photoelectrons will be emitted
  • All photoelectrons emitted will have kinetic energy equal to \(hν_0\)
  • Maximum kinetic energy of photoelectrons emitted can be \(hν_0\)
  • Maximum kinetic energy of photoelectrons emitted will be \(2hν_0\)
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The Correct Option is C

Solution and Explanation

According to Einstein’s photoelectric equation, the maximum kinetic energy \(K_{max}\) of the photoelectrons is given by:
\(K_{\text{max}} = h\nu - h\nu_0\)
where ν is the frequency of incident radiation and \(ν0 \)is the threshold frequency. If the radiation has a frequency \(2ν0\), then:
\(K_{\text{max}} = h(2\nu_0) - h\nu_0 = h\nu_0\)
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