Question

When a radiation of wavelength 3000 A° is incident on a metal, stopping potential is found to be 1.85 V and on making radiation of 4000 A° incident on it the stopping potential is found to be 0.82 V. Find (1) Planck's constant (2) Work function of the metal (3) Threshold wavelength of the metal..

Answer 1

Q can be obtained by integrating the above equation between the limits 0 and Q.

W = \int_{0}^{w}d

W = \int_{0}^{Q} (q/C).d

q = \frac{1}{C}\int_{0}^{Q}q.dq,

W = \frac{1}{C}[\frac{q^{2}}{2}]_{0}^{Q},

Or, W = 1/C[(Q2/2) – (0/2)], W = ½ (Q2/C)    …... (1) ,

If V = final potential of the capacitor, then Q = CV,

So, W = ½ [(CV)2/C],

Or,

W = ½ CV2 (2),

But, C = Q/V, Substituting for ‘C” in equation (1), we get,, W = ½ [Q2/(Q/V)]