Question

Scientists have made a light source whose spectral emissive power $\frac{dI}{d\lambda}$ is constant over visible range. Here I is intensity and $\lambda$ is wavelength .In other words, $\frac{dI}{d\lambda}$ graph is shown below. This beam of area $1m^2$ is incident on emitter plate of photoelectric tube. The collector plate is sufficiently positive so that tube is in saturation mode. Assume that each capable photon liberates 1 electron. If work function is 2eV, what is the current (in A) the tube? Round off to nearest integer. ($hc = 1240 \text{ eVnm}$)

Answer 1

3

Answer 2

The spectral emissive power is given by $\frac{dI}{d\lambda} = 0.031 \text{ W/m}^2 \times \text{nm}$ for $400 \text{ nm} \le \lambda \le 700 \text{ nm}$, and 0 otherwise. The work function is $\phi = 2 \text{ eV}$. The energy of a photon with wavelength $\lambda$ is $E = \frac{hc}{\lambda}$. For photoelectric emission, the photon energy must be greater than or equal to the work function: $E \ge \phi$. $\frac{hc}{\lambda} \ge \phi \implies \lambda \le \frac{hc}{\phi}$. Given $hc = 1240 \text{ eVnm}$ and $\phi = 2 \text{ eV}$, the maximum wavelength for photoelectric emission is $\lambda_{max} = \frac{1240 \text{ eVnm}}{2 \text{ eV}} = 620 \text{ nm}$. The light source emits photons in the range $400 \text{ nm} \le \lambda \le 700 \text{ nm}$. Therefore, the capable photons are those with wavelengths in the range $400 \text{ nm} \le \lambda \le 620 \text{ nm}$.

The intensity of capable photons is $I_{capable} = \int_{400 \text{ nm}}^{620 \text{ nm}} \frac{dI}{d\lambda} d\lambda = \int_{400}^{620} 0.031 d\lambda = 0.031 \times (620 - 400) = 0.031 \times 220 = 6.82 \text{ W/m}^2$. The total power of capable photons incident on the emitter plate is $P_{capable} = I_{capable} \times \text{Area} = 6.82 \text{ W/m}^2 \times 1 \text{ m}^2 = 6.82 \text{ W}$.

The number of photons with wavelengths between $\lambda$ and $\lambda + d\lambda$ incident per unit area per unit time is $dN' = \frac{dI}{d\lambda} \frac{d\lambda}{E} = \frac{dI}{d\lambda} \frac{d\lambda}{hc/\lambda} = \frac{dI}{d\lambda} \frac{\lambda}{hc} d\lambda$. The number of capable photons incident per second on the area $A$ is $N_{capable} = A \int_{400}^{620} \frac{dI}{d\lambda} \frac{\lambda}{hc} d\lambda$. Given $A = 1 \text{ m}^2$, $\frac{dI}{d\lambda} = 0.031 \text{ W/m}^2 \times \text{nm}$, $hc = 1240 \text{ eVnm}$. We need to convert W to J/s and eV to J. $1 \text{ W} = 1 \text{ J/s}$, $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$. $hc = 1240 \text{ eVnm} = 1240 \times 1.602 \times 10^{-19} \text{ Jnm} = 1.98648 \times 10^{-16} \text{ Jnm}$. $\frac{dI}{d\lambda} = 0.031 \text{ J/s/m}^2 \times \text{nm}$. $N_{capable} = (1 \text{ m}^2) \int_{400 \text{ nm}}^{620 \text{ nm}} (0.031 \text{ J/s/m}^2 \times \text{nm}) \frac{\lambda \text{ nm}}{1.98648 \times 10^{-16} \text{ Jnm}} d\lambda \text{ (in nm)}$. $N_{capable} = \frac{0.031}{1.98648 \times 10^{-16}} \int_{400}^{620} \lambda d\lambda \text{ s}^{-1}$. $\int_{400}^{620} \lambda d\lambda = [\frac{\lambda^2}{2}]_{400}^{620} = \frac{620^2 - 400^2}{2} = \frac{384400 - 160000}{2} = \frac{224400}{2} = 112200$. $N_{capable} = \frac{0.031}{1.98648 \times 10^{-16}} \times 112200 \approx 1.751 \times 10^{19} \text{ s}^{-1}$. Since each capable photon liberates 1 electron, the number of electrons emitted per second is equal to $N_{capable}$. The current is $I = N_{capable} \times e = (1.751 \times 10^{19} \text{ s}^{-1}) \times (1.602 \times 10^{-19} \text{ C})$. $I \approx 1.751 \times 1.602 \text{ A} \approx 2.805 \text{ A}$. Rounding off to the nearest integer, the current is $3 \text{ A}$.