When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is 6.0 V. This potential drops to 0.6 V if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively?

[Take $$\dfrac{hc}{e} = 1.24 \times 10^{-6}$$ J m C$$^{-1}$$.]

The photoelectric equation for a metal is
$$eV_s \;=\; \frac{hc}{\lambda}\;-\;\phi$$
where $$V_s$$ is the stopping potential, $$\lambda$$ is the wavelength of the incident light and $$\phi$$ is the work function of the metal.

First source (unknown wavelength $$\lambda$$) gives a stopping potential of $$V_{s1}=6.0\;{\rm V}$$.
$$e(6.0)=\frac{hc}{\lambda}-\phi \quad\;-(1)$$

Second source has wavelength $$\lambda_2 = 4\lambda$$ and produces a stopping potential of only $$V_{s2}=0.6\;{\rm V}$$. (Intensity does not affect the stopping potential.)
$$e(0.6)=\frac{hc}{4\lambda}-\phi \quad\;-(2)$$

Subtract $$(2)$$ from $$(1)$$ to eliminate $$\phi$$:

$$e(6.0-0.6)=\frac{hc}{\lambda}-\frac{hc}{4\lambda} =\frac{3hc}{4\lambda}$$

Simplify:

$$5.4e=\frac{3hc}{4\lambda}$$
$$\lambda=\frac{3hc}{4(5.4)e}=\frac{3}{21.6}\,\frac{hc}{e}$$

Given $$\dfrac{hc}{e}=1.24\times10^{-6}\;{\rm J\,m\,C^{-1}}$$, we obtain

$$\lambda =\frac{3}{21.6}\times1.24\times10^{-6}\;{\rm m} =1.72\times10^{-7}\;{\rm m}$$

Now substitute $$\lambda$$ in $$(1)$$ to find the work function. It is convenient to work in electron-volts by dividing energies by $$e$$:
Photon energy for the first source:
$$E_1=\frac{hc}{\lambda e} =\frac{1.24\times10^{-6}}{1.72\times10^{-7}} \;{\rm eV} \approx7.19\;{\rm eV}$$

Hence
$$\phi=E_1-6.0\;{\rm eV} =7.19-6.0 \approx1.19\;{\rm eV}\;\approx1.20\;{\rm eV}$$

Therefore, the wavelength of the first source and the work function of the metal are respectively
$$\boxed{1.72\times10^{-7}\;{\rm m},\;1.20\;{\rm eV}}$$

Option A is correct.