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Additional Exercises · 11.28

Q.A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used: λ1=3650 A˚,λ2=4047 A˚,λ3=4358 A˚,λ4=5461 A˚,λ5=6907 A˚\lambda_1 = 3650\ \text{Å}, \lambda_2 = 4047\ \text{Å}, \lambda_3 = 4358\ \text{Å}, \lambda_4 = 5461\ \text{Å}, \lambda_5 = 6907\ \text{Å}, The stopping voltages, respectively, were measured to be: V01=1.28 V,V02=0.95 V,V03=0.74 V,V04=0.16 V,V05=0 VV_{01} = 1.28\ \text{V}, V_{02} = 0.95\ \text{V}, V_{03} = 0.74\ \text{V}, V_{04} = 0.16\ \text{V}, V_{05} = 0\ \text{V}. Determine the value of Planck's constant hh, the threshold frequency and work function for the material.

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Einstein's equation eV0=hν−ϕeV_0=h\nu-\phi is a straight line when V0V_0 is plotted against ν\nu: slope =h/e=h/e, and the frequency-axis intercept is the threshold frequency ν0=ϕ/h\nu_0=\phi/h. Using two well-separated data points gives h≈6.57×10−34h\approx6.57\times10^{-34} J s, ν0≈5.1×1014\nu_0\approx5.1\times10^{14} Hz, and ϕ≈2.1\phi\approx2.1 eV.

Step 1 — Convert each wavelength to frequency (ν=c/λ\nu=c/\lambda).

ν1=3×1083650×10−10≈8.219×1014 Hz,ν4=3×1085461×10−10≈5.493×1014 Hz\nu_1 = \frac{3\times10^8}{3650\times10^{-10}} \approx 8.219\times10^{14}\ \text{Hz}, \qquad \nu_4 = \frac{3\times10^8}{5461\times10^{-10}} \approx 5.493\times10^{14}\ \text{Hz}

(similarly ν2≈7.413×1014\nu_2\approx7.413\times10^{14}, ν3≈6.884×1014\nu_3\approx6.884\times10^{14}, ν5≈4.343×1014\nu_5\approx4.343\times10^{14} Hz)

Step 2 — Find the slope using two well-separated points (λ1\lambda_1 and λ4\lambda_4).

Einstein's equation eV0=hν−ϕeV_0=h\nu-\phi means V0V_0 vs ν\nu is a straight line of slope h/eh/e:

slope=V01−V04ν1−ν4=1.28−0.168.219×1014−5.493×1014=1.122.726×1014≈4.11×10−15 V s\text{slope} = \frac{V_{01}-V_{04}}{\nu_1-\nu_4} = \frac{1.28-0.16}{8.219\times10^{14}-5.493\times10^{14}} = \frac{1.12}{2.726\times10^{14}} \approx 4.11\times10^{-15}\ \text{V s}

Step 3 — Planck's constant.

h=e×slope=(1.6×10−19)(4.11×10−15)≈6.57×10−34 J sh = e\times\text{slope} = (1.6\times10^{-19})(4.11\times10^{-15}) \approx 6.57\times10^{-34}\ \text{J s}

(within a couple of percent of the accepted value, consistent with this being real, rounded experimental data)

Step 4 — Threshold frequency and work function.

Using the line equation V0=(slope)ν+cV_0 = (\text{slope})\nu + c through point 1: 1.28=(4.11×10−15)(8.219×1014)+c⇒c≈−2.10 V1.28 = (4.11\times10^{-15})(8.219\times10^{14}) + c \Rightarrow c \approx -2.10\ \text{V}. …

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