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

Question

Choose correct option :
(A) Number of photons required for a light beam of 2000 pm wavelength and 1 Joule energy is \(1.01 \times 10^{16}\).
(B) Light with wavelength 300 nm has energy \(E_1\) and for wavelength 900 nm has energy \(E_2\), then \(E_1/E_2 = 1/3\).
(C) Frequency of light is \(4.5 \times 10^{16}\) Hz then its wavelength is \(6.7 \times 10^{-9}\) m.
(D) If electrons and protons are accelerated by same potential difference, then their de-Broglie wavelength are equal.

Answer 1

To address this problem, we must understand the photoelectric effect, a phenomenon where incident radiation on a metal surface ejects electrons, termed photoelectrons.

The relevant parameters are:

Threshold Frequency \((ν_0)\): The minimum frequency necessary for photoelectron emission.
Incident Radiation Frequency \((2ν_0)\): The incident frequency is double the threshold frequency, i.e., \(2ν_0\).
Incident Photon Energy: Calculated as \(h \times 2ν_0\), with \(h\) representing Planck's constant.
Work Function \((hν_0)\): The minimum energy required for electron emission, equivalent to the energy at the threshold frequency.

Einstein's photoelectric equation quantifies the maximum kinetic energy (\(K_{\text{max}}\)) of emitted photoelectrons as follows:

\[K_{\text{max}} = hν - hν_0\]

Upon substituting the provided frequencies:

\[K_{\text{max}} = h \times 2ν_0 - hν_0 = hν_0\]

This calculation indicates that the maximum kinetic energy of the emitted photoelectrons is \(hν_0\). Consequently, the accurate statement is:

"The maximum kinetic energy of emitted photoelectrons can be \(hν_0\)."

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

The Correct Option is C

Solution and Explanation

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