Dual Nature of Matter

JEE Main level, single correct, medium

169 of 200

Question 169 of 200

The stopping potential Vo{V}_{o} (in volt) as a function of frequency (v)(v) for a sodium emitter, is shown in the figure. The work function of sodium, form the data plotted in the figure, will be:
(Given: Planck’s constant (h)=6.63×10−34Js\left(h\right)=6.63\times {10}^{-34}Js , electron charge (e)(\text{e}) =1.6×10−19C)=1.6\times {10}^{-19}C)

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Answer: (D) 1.66eV1.66eV

From Einstein's photoelectric equation, hν=ϕ+eV0h\nu =\phi +e{V}_{0}.
⇒V0=hνe−ϕe......(1)\Rightarrow {V}_{0}=\dfrac{h\nu }{e}-\dfrac{\phi }{e}......\left(1\right)
At ν=4×1014Hz\nu =4\times {10}^{14}Hz, V0=0{V}_{0}=0.
Putting the value in equation (1)\left(1\right), we get
0=hve−ϕe0=\dfrac{hv}{e}-\dfrac{\phi }{e}
So, Work function (ϕ)=(hν)e(ineV)(\phi )=\dfrac{\left(h\nu \right)}{e} \left(in eV\right)
=6.63×10−34×(4×1014)1.6×10−19=\dfrac{6.63\times {10}^{-34}\times \left(4\times {10}^{14}\right)}{1.6\times {10}^{-19}}
=1.66eV=1.66 eV

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