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2.52    Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (b) Planck’s constant.

\\\lambda\ (\textup{nm})\qquad \qquad \qquad 500\quad 450\quad 400\\ \textup{v}\times 10^{-5} \ (\textup{cm s}^{-1})\quad \; 2.55\ \ 4.35\ \ 5.35

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We have the threshold wavelength.\lambda_{o} = 540\ nm

Then substituting this value in any of the equation ( look in the previous part), we get

\left ( \frac{hc}{10^{-9}} \right )\left ( \frac{1}{400}-\frac{1}{\lambda} \right )= \frac{1}{2}m(5.35\times10^3)^2

\left ( \frac{hc}{10^{-9}} \right )\left ( \frac{1}{400}-\frac{1}{540} \right )= \frac{1}{2}m(5.35\times10^3)^2

Taking the mass of an electron to be 9.11\times10^{-31}kg.

h= \frac{(9.11\times10^{-31})(5.35\times10^3)^2\times10^{-9}\times(400\times540)}{2\times(3.0\times10^8)(140)}

= 6.705\times10^{-34} Js  approximately.

Posted by

Divya Prakash Singh

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