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Question: Monochromatic radiation of wavelength \[640.2nm\] (\[1nm={{10}^{-9}}m\]) from a neon lamp irradiates...

Monochromatic radiation of wavelength 640.2nm640.2nm (1nm=10−9m1nm={{10}^{-9}}m) from a neon lamp irradiates photosensitive material made of cesium on tungsten. The stopping voltage is measured to be 0.54V0.54V. The source is replaced by an iron source and its 427.2nm427.2nm line irradiates the same photo-cell. Predict the new stopping voltage.

Explanation

Solution

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: eV0=hν−ϕ0e{{V}_{0}}=h\nu -{{\phi }_{0}}
or it can be written as, eV0=hcλ−ϕ0e{{V}_{0}}=\dfrac{hc}{\lambda }-{{\phi }_{0}}

Complete step by step answer: First of all we need to find the work function ϕ0{{\phi }_{0}}of the photosensitive material. For this, we rearrange the photoelectric energy relation as,
ϕ0=hcλ−eV0{{\phi }_{0}}=\dfrac{hc}{\lambda }-e{{V}_{0}}
Where, ϕ0{{\phi }_{0}} is the work function of photosensitive material.
hh is the planck's constant having a value of 6.626×10−346.626\times {{10}^{-34}}.
cc is the velocity of light.
λ\lambda is the wavelength from the neon lamp.
And V0{{V}_{0}} is the stopping voltage
The wavelength of the incident radiation from the neon lamp λ\lambda is given as 640.2nm640.2nm and the stopping voltage V0{{V}_{0}} is given as 0.54V0.54V.
To convert it into joules, we are multiplying V0{{V}_{0}} with ee.
Now, work function of photosensitive material is given by,
ϕ0=6.626×10−34×3×108640.2×10−9−1.6×10−19×0.54{{\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 nmnm to mm)
⇒ϕ0=3.0927×10−19−0.864×10−19\Rightarrow {{\phi }_{0}}=3.0927\times {{10}^{-19}}-0.864\times {{10}^{-19}}
ϕ0=2.2287×10−19{{\phi }_{0}}=2.2287\times {{10}^{-19}}
Therefore, the work function ϕ0{{\phi }_{0}} of photosensitive material is 2.2287×10−192.2287\times {{10}^{-19}}.
Now, to convert this to eVeVwe divide it with 1.6×10−191.6\times {{10}^{-19}}
⇒ϕ0=2.2287×10−191.6×10−19=1.39eV\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 427.2nm427.2nm.
i.e., stopping potential, eV0=hcλ−ϕ0e{{V}_{0}}=\dfrac{hc}{\lambda }-{{\phi }_{0}}
eV0=6.626×10−34×3×108427.2×109−2.2287×10−19e{{V}_{0}}=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{427.2\times {{10}^{9}}}-2.2287\times {{10}^{-19}}
eV0=2.401×10−19Je{{V}_{0}}=2.401\times {{10}^{-19}}J
To convert this into eVeV,
eV0=2.401×10−191.6×10−19=1.50eVe{{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 1.50eV1.50eV.

Note: We can also do this by only finding the threshold wavelength of the photosensitive material by using the equation, s.p=hc(1λ−1λ0)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 eVeV.