Question

The work function of a cathode is estimated using light of wavelength 310±1 nm. The stopping potential is measured to be 500±1mv. hce=1240vnm (known precisely). What is the error e (in %) in work function. Find value of 10e by rounding off to nearest integer.

Answer 1

4

Answer 2

The energy of the incident photon is given by $E = \frac{hc}{\lambda}$. The work function $\phi$ and the maximum kinetic energy of the emitted electron $K_{max}$ are related by Einstein's photoelectric equation: $E = \phi + K_{max}$. The maximum kinetic energy is related to the stopping potential $V_s$ by $K_{max} = eV_s$.

Thus, the work function is given by $\phi = E - K_{max} = \frac{hc}{\lambda} - eV_s$.

We are given $\frac{hc}{e} = 1240$ V nm. This implies $\frac{hc}{\lambda} = \frac{1240 \times e}{\lambda}$. To express energy in eV, we can write $\frac{hc}{e\lambda} = \frac{1240}{\lambda}$, where $\lambda$ is in nm. This gives the photon energy in eV. The stopping potential $V_s$ in Volts gives the maximum kinetic energy in eV directly ($K_{max, eV} = V_s$).

Given values: Wavelength $\lambda = 310$ nm, with uncertainty $\Delta \lambda = 1$ nm. Stopping potential $V_s = 500$ mV $= 0.500$ V, with uncertainty $\Delta V_s = 1$ mV $= 0.001$ V.

First, calculate the value of the work function $\phi$. Photon energy $E_{eV} = \frac{1240 \text{ eV nm}}{310 \text{ nm}} = 4$ eV. Maximum kinetic energy $K_{max, eV} = V_s = 0.500$ V $= 0.5$ eV. Work function $\phi = E_{eV} - K_{max, eV} = 4 \text{ eV} - 0.5 \text{ eV} = 3.5$ eV.

Next, calculate the error in the work function $\phi$. The work function is a function of $\lambda$ and $V_s$: $\phi(\lambda, V_s) = \frac{hc}{\lambda} - eV_s$. The absolute error in $\phi$ is given by $\Delta \phi = \left|\frac{\partial \phi}{\partial \lambda}\right|\Delta \lambda + \left|\frac{\partial \phi}{\partial V_s}\right|\Delta V_s$. $\frac{\partial \phi}{\partial \lambda} = -\frac{hc}{\lambda^2}$ $\frac{\partial \phi}{\partial V_s} = -e$ So, $\Delta \phi = \left|-\frac{hc}{\lambda^2}\right|\Delta \lambda + |-e|\Delta V_s = \frac{hc}{\lambda^2}\Delta \lambda + e\Delta V_s$.

To express $\Delta \phi$ in eV, we divide by $e$: $\Delta \phi_{eV} = \frac{\Delta \phi}{e} = \frac{hc}{e\lambda^2}\Delta \lambda + \Delta V_s$. Using $\frac{hc}{e} = 1240$ V nm: $\Delta \phi_{eV} = \frac{1240 \text{ V nm}}{\lambda^2 \text{ nm}^2}\Delta \lambda \text{ nm} + \Delta V_s \text{ V}$. $\Delta \phi_{eV} = \frac{1240}{\lambda^2}\Delta \lambda + \Delta V_s$ (in Volts, which is equivalent to eV for energy).

Substitute the values: $\lambda = 310$ nm, $\Delta \lambda = 1$ nm. $V_s = 0.5$ V, $\Delta V_s = 0.001$ V.

$\Delta \phi_{eV} = \frac{1240}{(310)^2} \times 1 + 0.001$ $\Delta \phi_{eV} = \frac{1240}{96100} + 0.001$ $\Delta \phi_{eV} = \frac{124}{9610} + \frac{1}{1000}$ $\Delta \phi_{eV} = \frac{124 \times 1000 + 9610 \times 1}{9610 \times 1000} = \frac{124000 + 9610}{9610000} = \frac{133610}{9610000} = \frac{13361}{961000}$ eV.

The percentage error in the work function is $e = \frac{\Delta \phi}{\phi} \times 100\%$. $e = \frac{13361/961000 \text{ eV}}{3.5 \text{ eV}} \times 100\%$ $e = \frac{13361}{961000} \times \frac{1}{3.5} \times 100$ $e = \frac{13361}{9610} \times \frac{1}{3.5}$ $e = \frac{13361}{9610 \times 3.5} = \frac{13361}{33635}$ $e \approx 0.39722\%$

We are asked to find the value of $10e$ by rounding off to the nearest integer. $10e \approx 10 \times 0.39722 = 3.9722$. Rounding off to the nearest integer, we get 4.

The final answer is $\boxed{4}$.

Explanation of the solution:

1. Calculate the work function $\phi$ using the formula $\phi = \frac{hc}{\lambda} - eV_s$ and the given values, converting to eV.
2. Determine the formula for the absolute error in $\phi$ using error propagation: $\Delta \phi = \left|\frac{\partial \phi}{\partial \lambda}\right|\Delta \lambda + \left|\frac{\partial \phi}{\partial V_s}\right|\Delta V_s$.
3. Calculate the partial derivatives and substitute them into the error formula.
4. Calculate the value of the absolute error $\Delta \phi$ in eV using the given uncertainties $\Delta \lambda$ and $\Delta V_s$.
5. Calculate the percentage error $e = \frac{\Delta \phi}{\phi} \times 100\%$.
6. Calculate $10e$ and round it to the nearest integer.

The final answer is $\boxed{4}$.