Question

When a light of wavelength '$\lambda$' falls on the emitter of a photocell, the maximum speed of emitted photoelectrons is '$V$'. If the incident wavelength is changed to $\frac{2\lambda}{3}$, the maximum speed of emitted photoelectrons will be

• A. less than $V \left(1.5\right)^{1/2}$
• B. equal to $V$
• C. greater than $V \left(1.5\right)^{1/2}$
• D. equal to $V \left(1.5\right)^{1/2}$

Hint:

When the incident photon energy increases by a certain factor (here, wavelength drops to $\frac{2}{3}$ so energy multiplies by $1.5$), the work function stays fixed. Since the work function doesn't scale up, the entire excess energy is transferred directly into the electron's kinetic energy. This causes the maximum kinetic energy to increase by a factor even larger than $1.5$, making the final speed strictly greater than $\sqrt{1.5}V$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem concerns the photoelectric effect. An incident light of wavelength $\lambda$ produces photoelectrons with a maximum speed $V$. We need to deduce the inequality boundary constraint for the new maximum speed $V'$ when the incident wavelength is shortened to $\frac{2\lambda}{3}$.

Step 2: Key Formula or Approach:
Apply Einstein's Photoelectric Equation relating incident photon energy to the metal work function ($\phi_0$) and the maximum kinetic energy of the ejected electrons: $$\frac{hc}{\lambda} = \phi_0 + \frac{1}{2}mV^2$$

Step 3: Detailed Explanation:
Let's write down the photoelectric equation for the first case with wavelength $\lambda$ and speed $V$: $$\frac{hc}{\lambda} = \phi_0 + \frac{1}{2}mV^2 \quad \text{--- (Equation 1)}$$ Now write down the equation for the second case with a shortened wavelength $\lambda' = \frac{2\lambda}{3}$ and a new speed $V'$: $$\frac{hc}{\left(\frac{2\lambda}{3}\right)} = \phi_0 + \frac{1}{2}mV'^2 \implies 1.5 \left(\frac{hc}{\lambda}\right) = \phi_0 + \frac{1}{2}mV'^2 \quad \text{--- (Equation 2)}$$ Let's substitute the value of $\frac{hc}{\lambda}$ from Equation 1 into Equation 2: $$1.5 \left(\phi_0 + \frac{1}{2}mV^2\right) = \phi_0 + \frac{1}{2}mV'^2$$ $$1.5\phi_0 + 1.5 \left(\frac{1}{2}mV^2\right) = \phi_0 + \frac{1}{2}mV'^2$$ Rearrange the equation to isolate the new kinetic energy term on one side: $$\frac{1}{2}mV'^2 = 1.5 \left(\frac{1}{2}mV^2\right) + 0.5\phi_0$$ Divide the entire equation by $\frac{1}{2}m$ to solve for the velocity squares: $$V'^2 = 1.5V^2 + \frac{\phi_0}{\frac{1}{2}m}$$ Since the work function $\phi_0$ and mass $m$ are strictly positive real quantities, the term $\frac{\phi_0}{\frac{1}{2}m} > 0$. Therefore, we can establish a strict inequality: $$V'^2 > 1.5V^2$$ Taking the square root on both sides: $$V' > V\left(1.5\right)^{1/2}$$

Step 4: Final Answer:
The new maximum speed will be greater than $V \left(1.5\right)^{1/2}$, which corresponds to option (C).