Question

In a photoemissive cell with exciting wavelength $λ$, the fastest electron has a speed $v$. If the exciting wavelength is changed to $\frac34λ$, then the speed of the fastest emitted electron will be

• A. $v\left(\frac34\right)¹/2$
• B. $v\left(\frac43\right)¹/2$
• C. less than $v\left(\frac43\right)¹/2$
• D. greater than $v\left(\frac43\right)¹/2$

Hint:

In photoelectric effect, $K_\max = \frachcλ - φ$, so speed does not scale directly with $1/λ$.

Answer 1

The Correct Option is C

Step 1: Einstein’s photoelectric equation: $K_\max = \frachcλ - φ$.

Step 2: For $λ' = \frac34λ$, $K'_\max = \frachc(3/4)λ - φ = \frac4hc3λ - φ$.

Step 3: Increase in kinetic energy is not directly proportional due to work function $φ$.

Step 4: Since $v \propto \sqrtK_\max}$, new speed is less than $v\sqrt\frac43}$.