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  • Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength

2.52    Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength

\\\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

Answers (1)

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Let us assume the threshold wavelength to be \lambda\ nm and the kinetic energy of the radiation is given as:

h(\nu-\nu_{o}) = \frac{1}{2}mv^2

hc\left ( \frac{1}{\lambda} - \frac{1}{\lambda_{o}} \right )= \frac{1}{2}mv^2

\Rightarrow hc\left ( \frac{1}{500\times10^9} - \frac{1}{\lambda\times10^{-9}} \right ) = \frac{1}{2}m\left ( 2.55\times10^5\times10^{-2} \right )^2

\left ( \frac{hc}{10^{-9}} \right )\left ( \frac{1}{500}-\frac{1}{\lambda} \right )= \frac{1}{2}m(2.55\times10^3)^2         .................................(1)

Similarly, we can also write,

\left ( \frac{hc}{10^{-9}} \right )\left ( \frac{1}{450}-\frac{1}{\lambda} \right )= \frac{1}{2}m(3.45\times10^3)^2        ...................................(2)

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

Now, dividing the equations (3) with (1),

\frac{\left ( \frac{\lambda -400}{400\lambda} \right )}{\left (\frac{\lambda -500}{500\lambda} \right )} = \frac{\left ( 5.35\times10^3 \right )^2}{\left ( 2.55\times10^3 \right )^2}

\frac{5\lambda -2000}{4\lambda -2000} = \frac{\left ( 5.35 \right )^2}{\left ( 2.55 \right )^2}

\frac{5\lambda -2000}{4\lambda -2000} = 4.40177

\Rightarrow 17.6070\lambda -5\lambda = 8803.537 -2000

\Rightarrow \lambda = \frac{6805.537}{12.607} = 539.8\ nm

Therefore, the wavelength is 540\ nm.

Posted by

Divya Prakash Singh

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