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Question

(C) The frequency of the $H_{\alpha}$ line of the Balmer series is given by $
u_0 = Rc(\frac{1}{2^2} - \frac{1}{3^2}) = Rc(\frac{1}{4} - \frac{1}{9})

Vedclass · Accepted Answer

$\frac{7 Rhc}{144}$

Vedclass · Answer

(C) The frequency of the $H_{\alpha}$ line of the Balmer series is given by $
u_0 = Rc(\frac{1}{2^2} - \frac{1}{3^2}) = Rc(\frac{1}{4} - \frac{1}{9}) = \frac{5Rc}{36}$. This is the threshold frequency $
u_0$ of the material.The frequency of the $H_{\beta}$ line of the Balmer series is given by $
u = Rc(\frac{1}{2^2} - \frac{1}{4^2}) = Rc(\frac{1}{4} - \frac{1}{16}) = \frac{3Rc}{16}$.According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = h
u - h
u_0$.Substituting the values:$K_{max} = h(\frac{3Rc}{16}) - h(\frac{5Rc}{36})$$K_{max} = Rhc(\frac{3}{16} - \frac{5}{36})$$K_{max} = Rhc(\frac{27 - 20}{144}) = \frac{7 Rhc}{144}$.

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