Question

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

• A. No photoelectrons will be emitted
• B. All photoelectrons emitted will have kinetic energy equal to \(hν_0\)
• C. Maximum kinetic energy of photoelectrons emitted can be \(hν_0\)
• D. Maximum kinetic energy of photoelectrons emitted will be \(2hν_0\)

Answer 1

The Correct Option is C

To solve this problem, we must understand the photoelectric effect, where radiation incident on a metallic surface can cause the emission of electrons, known as photoelectrons.

The key elements involved in this scenario are:

• Threshold Frequency \((ν_0)\): The minimum frequency required to emit photoelectrons.
• Incident Radiation Frequency \((2ν_0)\): In this case, the incident frequency is twice the threshold frequency, or \(2ν_0\).
• Energy of Incident Photon: Given by \(h \times 2ν_0\), where \(h\) is Planck's constant.
• Work Function \((hν_0)\): The minimum energy needed to emit an electron, corresponding to the threshold frequency.

According to Einstein's photoelectric equation, the maximum kinetic energy (\(K_{\text{max}}\)) of emitted photoelectrons is calculated by:

\[K_{\text{max}} = hν - hν_0\]

Substituting the given frequencies:

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

This calculation shows that the maximum kinetic energy of the emitted photoelectrons is \(hν_0\). Therefore, the correct statement is:

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