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NCERT Exemplar · Q21

Q.In a photoelectric experiment carried out with two different metals A and B, the stopping potential VstopV_{stop} (in volts) is measured as a function of the frequency ν\nu (in Hz) of the incident light. For each metal the measured points lie on a straight line, and the two straight lines are parallel to each other (they have the same slope). The line for metal A lies to the left of the line for metal B. Extrapolated down to zero stopping potential, the line for A meets the frequency axis at a threshold frequency of about 5×10145\times10^{14} Hz, while the line for B meets it at about 10×101410\times10^{14} Hz; each line then rises with the same slope (for instance, line A passes close to the point ν=10×1014\nu=10\times10^{14} Hz, Vstop=2V_{stop}=2 V, and line B passes close to the point ν=15×1014\nu=15\times10^{14} Hz, Vstop=2V_{stop}=2 V).

(i) Which material, A or B, has the higher work function?
(ii) Given that the electric charge of an electron is 1.6×10−191.6\times10^{-19} C, find the value of Planck's constant hh obtained from this experiment for both A and B, and comment on whether the result is consistent with Einstein's photoelectric theory.
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Both metals give parallel straight lines of VstopV_{stop} versus ν\nu, so they share the same slope h/eh/e. Metal B's line meets the frequency axis at the higher threshold frequency, so B has the larger work function. The common slope gives h≈6.4×10−34h\approx6.4\times10^{-34} J·s for both metals, matching Einstein's constant — confirming his theory.

Concept — Einstein's photoelectric equation

The stopping potential satisfies

eVstop=hν−ϕ0,eV_{stop}=h\nu-\phi_0,

which rearranges to

Vstop=he ν−ϕ0e.V_{stop}=\frac{h}{e}\,\nu-\frac{\phi_0}{e}.

This is a straight line in the (ν, Vstop)(\nu,\,V_{stop}) plane with slope he\dfrac{h}{e} and a frequency-axis intercept (where Vstop=0V_{stop}=0) at the threshold frequency ν0=ϕ0h\nu_0=\dfrac{\phi_0}{h}.

(i) Which has the higher work function

The work function is ϕ0=hν0\phi_0=h\nu_0, so a larger threshold frequency (an intercept further to the right on the frequency axis) means a larger work function. Line B meets the axis at ν0B≈10×1014\nu_{0B}\approx10\times10^{14} Hz while line A meets it at ν0A≈5×1014\nu_{0A}\approx5\times10^{14} Hz, so

ϕ0B=hν0B>ϕ0A=hν0A.\phi_{0B}=h\nu_{0B}>\phi_{0A}=h\nu_{0A}.

Metal B has the higher work function.

(ii) Value of Planck's constant

The slope of each line is h/eh/e, so h=e×(slope)h=e\times(\text{slope}).

For material A, using Vstop=0V_{stop}=0 at ν=5×1014\nu=5\times10^{14} Hz and Vstop=2V_{stop}=2 V at ν=10×1014\nu=10\times10^{14} Hz:

slopeA=2−0(10−5)×1014=4×10−15 V⋅s,\text{slope}_A=\frac{2-0}{(10-5)\times10^{14}}=4\times10^{-15}\ \text{V·s},

hA=e⋅slopeA=(1.6×10−19)(4×10−15)=6.4×10−34 J⋅s.h_A=e\cdot\text{slope}_A=(1.6\times10^{-19})(4\times10^{-15})=6.4\times10^{-34}\ \text{J·s}.

For material B, using Vstop=0V_{stop}=0 at ν=10×1014\nu=10\times10^{14} Hz and Vstop=2V_{stop}=2 V at ν=15×1014\nu=15\times10^{14} Hz:

slopeB=2−0(15−10)×1014=4×10−15 V⋅s,\text{slope}_B=\frac{2-0}{(15-10)\times10^{14}}=4\times10^{-15}\ \text{V·s},

hB=e⋅slopeB=(1.6×10−19)(4×10−15)=6.4×10−34 J⋅s.h_B=e\cdot\text{slope}_B=(1.6\times10^{-19})(4\times10^{-15})=6.4\times10^{-34}\ \text{J·s}.

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