JEE Advance - Physics (2009 - Paper 2 Offline - No. 2)
Photoelectric effect experiments are performed using three different metal plates p, q and r having work functions $$\phi_p=2.0~\mathrm{eV}$$, $$\phi_q=2.5~\mathrm{eV}$$ and $$\phi_r=3.0~\mathrm{eV}$$, respecticely. A light beam containing wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is (Take hc = 1240 eV nm)
Explanation
We have
$$E = \left( {{{hc} \over \lambda }} \right)$$ J $$ \Rightarrow E = \left( {{{1240} \over {\lambda \,nm}}} \right)$$ eV
Therefore,
$$\lambda_1$$ = 550 nm, $$E_1$$ = 2.25 eV
$$\lambda_2$$ = 450 nm, $$E_2$$ = 2.75 eV
$$\lambda_3$$ = 350 nm, $$E_3$$ = 3.5 eV
Also,
$$\phi_p=2$$ eV, all $$\lambda$$'s cause emissions.
$$\phi_q=2.5$$ eV, last two $$\lambda$$'s cause emissions.
$$\phi_r=3$$ eV, only the last $$\lambda$$ causes emissions.
That is, $$I_p > I_q > I_r$$.
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