Question

A photoelectric surface is illuminated successively by monochromatic light of wavelength ‘(λ)’ and ‘($\frac {λ}{2}$)'. If the maximum kinetic energy of the emitted photoelectrons in the first case is one-third that in the second case, the work function of the surface of the materialis (c = speed of light, h = Planck’s constant.)

• A. $\frac {hc}{3λ}$
• B. $\frac {hc}{2λ}$
• C. $\frac {2hc}{λ}$
• D. $\frac {hc}{λ}$

Answer 1

Answer: (B)

$\frac {hc}{2λ}$

Answer 2

Let's denote the work function of the surface of the material by φ. The maximum kinetic energy (KEmax) of the emitted photoelectrons can be determined using the equation:
KEmax = hν - φ
In the first case, the incident light has a wavelength of λ. The corresponding frequency is given by ν = $\frac {c}{λ}$,
KEmax1 = h($\frac {c}{λ}$) - φ
In the second case, the incident light has a wavelength of $\frac {λ}{2}$. The corresponding frequency is ν = $\frac {c}{(λ/2)}$ = $\frac {2c}{λ}$
Thus, the maximum kinetic energy (KEmax2) in the second case can be expressed as:
KEmax2 = h($\frac {2c}{λ}$) - φ
Given that KEmax1 is one-third of KEmax2, we can write the equation:
KEmax1 = $\frac {1}{3}$KEmax2
Substituting the expressions for KEmax1 and KEmax2, we have:
h($\frac {c}{λ}$) - φ = $\frac {1}{3}$[h($\frac {2c}{λ}$) - φ]
Simplifying this equation yields:
h($\frac {c}{λ}$) - φ = $\frac {4}{3}$(h($\frac {2c}{λ}$)) - $\frac {1}{3}$φ h($\frac {c}{λ}$) - φ = $\frac {4}{3}$(h($\frac {c}{λ}$)) - $\frac {1}{3}$φ
Rearranging the equation, we find:
h($\frac {c}{λ}$) - $\frac {4}{3}$(h($\frac {c}{λ}$)) = φ - $\frac {1}{3}$φ
Multiplying through by 3, we get:
3h($\frac {c}{λ}$) - 4h($\frac {c}{λ}$) = 3φ - φ h($\frac {c}{λ}$) = 2φ
Rearranging further, we have:
φ = -$\frac {{h(\frac {c}{λ})}}{2}$
Simplifying this expression, we find: φ = -$\frac {hc}{2λ}$
Therefore, the work function of the surface of the material is given by option (B) $\frac {hc}{2λ}$.