Question

Monochromatic radiation of wavelength \[640.2nm\] (\[1nm={{10}^{-9}}m\]) from a neon lamp irradiates photosensitive material made of cesium on tungsten. The stopping voltage is measured to be \[0.54V\]. The source is replaced by an iron source and its \[427.2nm\] line irradiates the same photo-cell. Predict the new stopping voltage.

Answer 1

Hint: To solve this question we use the photoelectric energy relation. First we find the work function of the photosensitive material and then substitute it in the next step to find the stopping voltage of the iron source.

Formula used: \[e{{V}_{0}}=h\nu -{{\phi }_{0}}\]
or it can be written as, \[e{{V}_{0}}=\dfrac{hc}{\lambda }-{{\phi }_{0}}\]

Complete step by step answer: First of all we need to find the work function \[{{\phi }_{0}}\]of the photosensitive material. For this, we rearrange the photoelectric energy relation as,
 \[{{\phi }_{0}}=\dfrac{hc}{\lambda }-e{{V}_{0}}\]
Where, \[{{\phi }_{0}}\] is the work function of photosensitive material.
\[h\] is the planck's constant having a value of \[6.626\times {{10}^{-34}}\].
 \[c\] is the velocity of light.
\[\lambda \] is the wavelength from the neon lamp.
And \[{{V}_{0}}\] is the stopping voltage
The wavelength of the incident radiation from the neon lamp \[\lambda \] is given as \[640.2nm\] and the stopping voltage \[{{V}_{0}}\] is given as \[0.54V\].
To convert it into joules, we are multiplying \[{{V}_{0}}\] with \[e\].
Now, work function of photosensitive material is given by,
\[{{\phi }_{0}}=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{640.2\times {{10}^{-9}}}-1.6\times {{10}^{-19}}\times 0.54\] -- (we converted \[nm\] to \[m\])
\[\Rightarrow {{\phi }_{0}}=3.0927\times {{10}^{-19}}-0.864\times {{10}^{-19}}\]
\[{{\phi }_{0}}=2.2287\times {{10}^{-19}}\]
Therefore, the work function \[{{\phi }_{0}}\] of photosensitive material is \[2.2287\times {{10}^{-19}}\].
Now, to convert this to \[eV\]we divide it with \[1.6\times {{10}^{-19}}\]
 \[\Rightarrow {{\phi }_{0}}=\dfrac{2.2287\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}=1.39eV\]
Now we need to find the stopping potential when the neon source is substituted with the iron source of \[427.2nm\].
i.e., stopping potential, \[e{{V}_{0}}=\dfrac{hc}{\lambda }-{{\phi }_{0}}\]
\[e{{V}_{0}}=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{427.2\times {{10}^{9}}}-2.2287\times {{10}^{-19}}\]
\[e{{V}_{0}}=2.401\times {{10}^{-19}}J\]
To convert this into \[eV\],
 \[e{{V}_{0}}=\dfrac{2.401\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}=1.50eV\]

Therefore, the stopping potential of iron source is found to be \[1.50eV\].

Note: We can also do this by only finding the threshold wavelength of the photosensitive material by using the equation, \[s.p=hc\left( \dfrac{1}{\lambda }-\dfrac{1}{{{\lambda }_{0}}} \right)\] and by substituting the threshold wavelength in the next step. One thing to be careful in this type of questions are the conversions to \[eV\].