Question

Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and (b) Planck's constant.

$\lambda $ (nm)	500	450	400
v$\times {{10}^{-5}}$ (cm/sec)	2.55	4.35	5.35

Answer 1

Hint:. The formula to calculate the kinetic energy of the radiation is as follows.
\[h(v-{{v}_{o}})=\dfrac{1}{2}m{{v}^{2}}\]
h = Planck's constant, m = mass of the wave, v = speed of the wave

Complete step by step answer:
- In the question it is given that sodium metal irradiated with different wavelengths have different frequencies and we have to calculate the threshold wavelength and Planck's constant from the given data.

(a) First we will calculate the threshold wavelength for the sodium metal from the given data.
- Assume that threshold wavelength is ${{\text{ }\!\!\lambda\!\!\text{ }}_{\text{o}}}\text{ nm or }{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{o}}}\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-9}}}\text{ m}$ .
- Now substitute the wavelength and frequency given in the question in the following formula.
\[\begin{align}
  & h(v-{{v}_{o}})=\dfrac{1}{2}m{{v}^{2}} \\
 & hc(\dfrac{1}{\lambda }-\dfrac{1}{{{\lambda }_{o}}})=\dfrac{1}{2}m{{v}^{2}} \\
 & hc\left( \dfrac{1}{500\times {{10}^{-9}}}-\dfrac{1}{{{\lambda }_{o}}\times {{10}^{-9}}} \right)=\dfrac{1}{2}m{{\left( 2.55\times {{10}^{5}} \right)}^{2}}\to (1) \\
\end{align}\]

- Likewise substitute the remaining values
\[hc\left( \dfrac{1}{450\times {{10}^{-9}}}-\dfrac{1}{{{\lambda }_{o}}\times {{10}^{-9}}} \right)=\dfrac{1}{2}m{{\left( 4.35\times {{10}^{5}} \right)}^{2}}\to (2)\]
\[hc\left( \dfrac{1}{400\times {{10}^{-9}}}-\dfrac{1}{{{\lambda }_{o}}\times {{10}^{-9}}} \right)=\dfrac{1}{2}m{{\left( 5.32\times {{10}^{5}} \right)}^{2}}\to (3)\]
- Now we have three equations numbered as I, 2 and 3,
- To get the threshold wavelength divide equation 2 with equation 1.
\[\begin{align}
  & \dfrac{hc\left( \dfrac{1}{450\times {{10}^{-9}}}-\dfrac{1}{{{\lambda }_{o}}\times {{10}^{-9}}} \right)}{hc\left( \dfrac{1}{500\times {{10}^{-9}}}-\dfrac{1}{{{\lambda }_{o}}\times {{10}^{-9}}} \right)}=\dfrac{\dfrac{1}{2}m{{\left( 4.35\times {{10}^{5}} \right)}^{2}}}{\dfrac{1}{2}m{{\left( 2.55\times {{10}^{5}} \right)}^{2}}} \\
 & \dfrac{{{\lambda }_{o}}-450}{450{{\lambda }_{o}}}\times \dfrac{500{{\lambda }_{o}}}{{{\lambda }_{o}}-500}={{\left( \dfrac{4.35}{2.55} \right)}^{2}} \\
 & \dfrac{{{\lambda }_{o}}-450}{{{\lambda }_{o}}-500}=2.91 \\
 & {{\lambda }_{o}}=526 nm. \\
\end{align}\]
- Therefore the threshold wavelength ${{\lambda }_{o}}$ = 526 nm.

(b) - Now substitute the value of threshold wavelength we have got form the above calculation in equation 3 to get the value of Planck's constant.
\[\begin{align}
  & hc\left( \dfrac{1}{400\times {{10}^{-9}}}-\dfrac{1}{{{\lambda }_{o}}\times {{10}^{-9}}} \right)=\dfrac{1}{2}m{{\left( 5.32\times {{10}^{5}} \right)}^{2}}\to (3) \\
 & h\times 3\times {{10}^{8}}\left( \dfrac{1}{400\times {{10}^{-9}}}-\dfrac{1}{526\times {{10}^{-9}}} \right)=\dfrac{1}{2}(9.11\times {{10}^{-31}}){{\left( 5.32\times {{10}^{5}} \right)}^{2}} \\
 & h=6.84\times {{10}^{-34}}Js \\
\end{align}\]
- Therefore the Planck's constant value is $h=6.84\times {{10}^{-34}}$ Js.

Note: Threshold wavelength is nothing but the maximum wavelength or minimum frequency of the incident radiation required to eject photons from the given surface. Here in the question the given surface is sodium metal to eject photons from the sodium surface the required threshold wavelength is 526 nm.