Question 8: Dual Nature of Matter and Radiation

Question

The stopping potential when a metal surface is illuminated by light of wavelength $\lambda$ is 15 V . The stopping potential when the same surface is illuminated by light of wavelength $4 \lambda$ is 3 V . The ratio of threshold wavelength to the initial incident wavelength $\lambda$ is:

Accepted Answer

Let $A=\dfrac{hc}{e\,\lambda}$ (in eV) and let $\phi_e$ be the work-function in eV.  The photoelectric equations for stopping potentials $V_{1}=15\,$V at $\lambda$ and $V_{2}=3\,$V at $4\lambda$ are
$V_{1}=A-\phi_e\quad\Longrightarrow\quad 15=A-\phi_e,$
$V_{2}=\dfrac{A}{4}-\phi_e\quad\Longrightarrow\quad 3=	frac{A}{4}-\phi_e.$
Subtracting (2) from (1):
$$15-3=\Bigl(A-	frac{A}{4}\Bigr)\quad\Longrightarrow\quad 12=	frac{3A}{4} \quad\Longrightarrow\quad A=16	ext{ eV}.$$
Then from (1), $\phi_e=A-15=1\,$eV.
The threshold wavelength $\lambda_0$ satisfies
$$\phi_e=\frac{hc}{e\,\lambda_0}\quad\Longrightarrow\quad \lambda_0=\frac{hc}{e\,\phi_e}=\frac{A\,\lambda}{\phi_e}=16\,\lambda.$$
So the ratio $\lambda_0:\lambda$ is $\boxed{16:1}$ (Option C).

Answer

$1: 16$

Answer

$1: 15$

Answer

$16: 1$

Answer

$15: 1$

Dual Nature of Matter and Radiation

The stopping potential when a metal surface is illuminated by light of wavelength \lambda is 15 V . The stopping potential when the same surface is illuminated by light of wavelength 4 \lambda is 3 V . The ratio of threshold wavelength to the initial incident wavelength \lambda is:

Choose the correct answer: