JEE Advance - Physics (2016 - Paper 1 Offline - No. 7)
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 (V0) are given below:
Given that c = 3 $$ \times $$ 108 ms-1 and e = 1.6 $$ \times $$ 10-19 C, Planck's constant (in units of J-s) found from such an experiment is) :
| $$\lambda \left( {\mu m} \right)$$ | V0(Volt) |
|---|---|
| 0.3 | 2.0 |
| 0.4 | 1.0 |
| 0.5 | 0.4 |
Given that c = 3 $$ \times $$ 108 ms-1 and e = 1.6 $$ \times $$ 10-19 C, Planck's constant (in units of J-s) found from such an experiment is) :
6.0 $$ \times $$ 10-34
6.6 $$ \times $$ 10-34
6.4 $$ \times $$ 10-34
6.8 $$ \times $$ 10-34
Explanation
$${{hc} \over \lambda } - \phi = e{V_0}$$ ($$\phi$$ = work function)
$${{hc} \over {0.3 \times {{10}^{ - 6}}}} - \phi = 2e$$ .... (i)
$${{hc} \over {0.4 \times {{10}^{ - 6}}}} - \phi = 1e$$ .... (ii)
Subtracting Eq. (ii) from Eq. (i)
$$hc\left( {{1 \over {0.3}} - {1 \over {0.4}}} \right){10^6} = e$$
$$hc\left( {{{0.1} \over {0.12}} \times {{10}^6}} \right) = e$$
$$h = 0.64 \times {10^{ - 33}}$$
$$ = 6.4 \times {10^{ - 34}}$$ J-s
$${{hc} \over {0.3 \times {{10}^{ - 6}}}} - \phi = 2e$$ .... (i)
$${{hc} \over {0.4 \times {{10}^{ - 6}}}} - \phi = 1e$$ .... (ii)
Subtracting Eq. (ii) from Eq. (i)
$$hc\left( {{1 \over {0.3}} - {1 \over {0.4}}} \right){10^6} = e$$
$$hc\left( {{{0.1} \over {0.12}} \times {{10}^6}} \right) = e$$
$$h = 0.64 \times {10^{ - 33}}$$
$$ = 6.4 \times {10^{ - 34}}$$ J-s
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