JEE Advanced 2016 · Paper 1 · Physics Q1

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In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength $(\lambda)$ of incident light and the corresponding stopping potential $(V_0)$ are given below:

$\lambda$ (μm)	$V_0$ (Volt)
0.3	2.0
0.4	1.0
0.5	0.4

Given that $c = 3\times10^{8}\ \mathrm{m\,s^{-1}}$ and $e = 1.6\times10^{-19}$ C, Planck's constant (in units of J s) found from such an experiment is

6.0\times10^{-34}

6.4\times10^{-34}

6.6\times10^{-34}

6.8\times10^{-34}

Answer and worked solution

Official answer: B

Einstein's equation. $eV_0=\dfrac{hc}{\lambda}-\phi$ , so two data points eliminate $\phi$ :
$e(V_1-V_2)=hc\left(\frac1{\lambda_1}-\frac1{\lambda_2}\right).$

Use 0.3 μm and 0.4 μm.
$1.6\times10^{-19}\times1.0=h\times3\times10^8\times\left(\frac1{0.3}-\frac1{0.4}\right)\times10^6=h\times2.5\times10^{14}.$
So $h=6.4\times10^{-34}$ J s. The pair 0.4 μm, 0.5 μm gives the same value.

(B)

Topics:Photoelectric effect

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