Final answer:
Nickel will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell.
Step-by-step explanation:
The work function of a metal determines whether or not it will emit photoelectrons when exposed to radiation. The work function is the minimum amount of energy required to remove an electron from the surface of a metal. In this case, the work functions of the metals are given as follows: Na: 2.75 eV, K: 2.30 eV, Mo: 4.17 eV, Ni: 5.15 eV.
To determine which metal will not give photoelectric emission for a radiation of wavelength 3300 Å (3300 x 10^-10 m) from a He-Cd laser placed 1 m away from the photocell, we can use the formula:
E = hc/λ - Φ
where E is the energy of the photon, h is the Planck's constant (6.63 x 10^-34 J s), c is the speed of light (3 x 10^8 m/s), λ is the wavelength, and Φ is the work function.
For Na: E = (6.63 x 10^-34 J s)(3 x 10^8 m/s) / (3300 x 10^-10 m) - 2.75 eV
For K: E = (6.63 x 10^-34 J s)(3 x 10^8 m/s) / (3300 x 10^-10 m) - 2.30 eV
For Mo: E = (6.63 x 10^-34 J s)(3 x 10^8 m/s) / (3300 x 10^-10 m) - 4.17 eV
For Ni: E = (6.63 x 10^-34 J s)(3 x 10^8 m/s) / (3300 x 10^-10 m) - 5.15 eV
Calculating each of these values, we find that for Na: E = 4.78 eV, for K: E = 5.02 eV, for Mo: E = 3.20 eV, and for Ni: E = 2.73 eV.
From these calculations, we can see that the metal Ni (nickel) will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell, as the energy of the photon is lower than the work function of Ni.