NEETPhysicsDual Nature of Matter
The variation of stopping potential ( V₀ ) versus the reciprocal of incident wavelength ( 1/ ) is plotted for two different metals P and Q. The resulting graphs are parallel straight lines. The x-intercept of the line for metal P is greater than the x-intercept of the line for metal Q ( x_P > x_Q ). Monochromatic light of wavelength is incident on both metals such that the value of 1/ lies exactly halfway between x_Q
Options
- AMetal P emits photoelectrons, metal Q does not emit, and _P > _Q .
- BMetal P emits photoelectrons, metal Q does not emit, and _P < _Q .
- CMetal Q emits photoelectrons, metal P does not emit, and _P < _Q .
- DMetal Q emits photoelectrons, metal P does not emit, and _P > _Q .
Correct answer
D. Metal Q emits photoelectrons, metal P does not emit, and _P > _Q .
Step-by-step solution
According to Einstein's photoelectric equation: eV₀ = hc - V₀ = ( hc e ) 1 - e In a graph of V₀ versus 1/ , the x-intercept (where V₀ = 0 ) is given by x = hc . Since x_P > x_Q , it follows that _P hc > _Q hc , which means _P > _Q . The incident light has a value of 1/ that lies exactly halfway between x_Q and x_P . Thus, x_Q For metal Q, 1 > x_Q 1 > _Q hc hc > _Q . Since the incident photon energy is greater than the work function, metal Q emits photoelectrons. For metal P, 1 Therefore, metal Q emits photoelectron