Question

The light of wavelength '$\lambda$' is incident on the surface of metal of work function $\phi$ and emits the electron. The maximum velocity of electron emitted is [$m = \text{mass of electron and } h = \text{Planck's constant, } c = \text{velocity of light}$]

• A. $\left[ \frac{2(hc - \lambda)}{m\lambda} \right]^{\frac{1}{2}}$
• B. $\left[ \frac{2(hc - \phi)\lambda}{mc} \right]$
• C. $\left[ \frac{2(hc - \lambda)}{m\lambda} \right]$
• D. $\left[ \frac{2(hc - \lambda\phi)}{m\lambda} \right]^{\frac{1}{2}}$

Hint:

To quickly verify your algebraic steps, check the dimensions or units of the terms inside the parentheses. Since $hc/\lambda$ represents energy, the product $\lambda\phi$ ensures that both terms in the numerator have matching units of (Energy $\times$ Length), which makes the expression dimensionally correct.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Monochromatic light with a wavelength $\lambda$ strikes a metal surface that has a work function $\phi$, causing photoelectric emission. We need to isolate and express the maximum velocity $v_{\text{max}}$ of the ejected photoelectrons.

Step 2: Key Formula or Approach:
According to Einstein's Photoelectric Equation, the maximum kinetic energy of an emitted electron is equal to the energy of the incident photon minus the work function of the metal surface: $$K.E._{\text{max}} = \frac{1}{2}mv_{\text{max}}^2 = E - \phi$$ The energy of a photon in terms of its wavelength is $E = \frac{hc}{\lambda}$. Substituting this into the energy balance equation allows us to solve for $v_{\text{max}}$.

Step 3: Detailed Explanation:
Write out the photoelectric energy balance equation: $$\frac{1}{2}mv_{\text{max}}^2 = \frac{hc}{\lambda} - \phi$$ Find a common denominator for the terms on the right-hand side to merge them into a single fraction: $$\frac{1}{2}mv_{\text{max}}^2 = \frac{hc - \lambda\phi}{\lambda}$$ Multiply both sides of the equation by 2 and divide by the mass parameter $m$ to isolate $v_{\text{max}}^2$: $$v_{\text{max}}^2 = \frac{2(hc - \lambda\phi)}{m\lambda}$$ Take the square root of both sides to get the expression for the maximum velocity: $$v_{\text{max}} = \left[ \frac{2(hc - \lambda\phi)}{m\lambda} \right]^{\frac{1}{2}}$$ This matches option (D).

Step 4: Final Answer:
The maximum velocity of the emitted electron is $\left[ \frac{2(hc - \lambda\phi)}{m\lambda} \right]^{\frac{1}{2}}$, which corresponds to option (D).