Consider white light whose wavelength spread is from $400nm$ to $700nm$. Its energy is uniformly distributed in this spectrum. The light is incident on metal A of work function $1.55eV$. Saturation photocurrent is $6mA$. Now the same light is incident on metal B, work function $2.48eV$. Calculate the total energy that was being carried by the photons that eject photoelectrons and also calculate the saturation photocurrent while experimenting with metal B. [Take $hc = 1240 \space eV\space nm$]

I am actually struggling with the first part itself which apparently has nothing to do with photoelectric effect (maybe a little). The second part appears to be more difficult (I have no idea how to start off).

Here is what've done so far.

Attempt to solve part 1 (total energy associated with the band of wavelengths)

Let $n_\lambda$ be the number of photons that are produced for a selected wavelength $\lambda$. The question says that the energy is distributed uniformly which can be mathematically written as,

$$E = h\nu =h\frac{c}{\lambda}$$ where the symbols carry their usual meanings.

$$n_\lambda h\frac{c}{\lambda} = constant$$

Energy is uniformly distributed so the number of photons for the selected wavelength times the energy associated with the photon should be a constant.

Differentiating both sides we get, $$hc . d(\frac{n_\lambda}{\lambda}) = 0$$ $$\lambda .dn_\lambda + n_\lambda. d\lambda = 0$$

I don't know how to proceed from here.

$$E_\lambda = n_\lambda h\frac{c}{\lambda}$$

I initially thought that I could obtain a differential equation from the above equation and integrate it with appropriate limits to obtain the total energy but that just doesn't work for obvious reasons (already used that equation once to get a constraint; relationship b/w $\lambda$ and $n_\lambda$)

Please give some hints and suggest ideas on how to continue with the problem. You may also give new methods to approach the problem.

Second Part (saturation current)

I have no clue how to calculate the saturation current. Need hints to get started.

Useful information which I have already gathered:

  • Only 400nm to 500nm range can eject photoelectrons for metal B (work function limitations)

If you will, comment some hints before posting the complete answer so that I can work it out myself.

Would appreciate if someone could come up with a better title.

  • $\begingroup$ You know that if you integrate from 400nm to the cutoff frequency, you'll get 6mA. The cutoff where the energy of the photon is equal to the work function. Find out how many electrons per second is 6mA and that is how many photons. $\endgroup$ – Jonathan Wheeler May 1 '16 at 18:20
  • $\begingroup$ 6mA is for metal A. The question asks to find total energy and saturation current for metal B. The number of photons varies according to wavelength so as to keep energy uniformly distributed. Only 400nm to 500nm range of wavelength is capable to release photoelectrons from metal B whereas all the photons can release photoelectrons for metal A. $\endgroup$ – Yashas May 2 '16 at 3:34
  • $\begingroup$ An important step in solving this problem is the intensity of light/number of photons, which is constant, across these two metals. You call this n_\lambda $\endgroup$ – Jonathan Wheeler May 2 '16 at 3:49
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    $\begingroup$ You figured out what $n_\lambda$ is based on the 6mA? $\endgroup$ – Jonathan Wheeler May 2 '16 at 4:28
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    $\begingroup$ I think this should do : Let $P$ be the power of each wavelength of in the given spectrum. $$N=\frac{P\lambda}{hc}$$ $$\implies dN=\frac{P d\lambda}{hc} $$ $$\implies \int_{0}^{I/e} dN=\int_{400 \space nm}^{700 \space nm}\frac{P d\lambda}{hc}$$ Find $P$. Using this $P$ solve the next part of the problem which asks for total power of spectrum which can cause photo electric effect $$P_{T}= \int_{400 nm}^{\lambda_o nm} P d{\lambda}$$ $\lambda_o$ is threshold wavelength. (I've used power instead of energy because no time limit has been specified.) $\endgroup$ – user102705 Feb 22 '17 at 10:12

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