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Step-by-Step Solution
Step 1: Write down the photoelectric equation for each wavelength
When radiation of wavelength $ \lambda $ strikes the metal, the stopping potential $V_{s1}$ is given by:
$ V_{s1} = \frac{hc}{\lambda} - \phi \quad \dots (1)$
When the wavelength is doubled ($2\lambda$), the new stopping potential $V_{s2}$ is:
$ V_{s2} = \frac{hc}{2\lambda} - \phi \quad \dots (2)$
Here, $h$ is Planck’s constant, $c$ is the speed of light, and $ \phi $ is the work function (in suitable units, such that the potential is directly in volts).
Step 2: Substitute the given values for the stopping potentials
We are given:
$ V_{s1} = 4.8 \text{ V}$ for wavelength $ \lambda $
$ V_{s2} = 1.6 \text{ V}$ for wavelength $ 2\lambda $
So, the equations become:
$ 4.8 = \frac{hc}{\lambda} - \phi \quad \dots (1)$
$ 1.6 = \frac{hc}{2\lambda} - \phi \quad \dots (2)$
Step 3: Subtract the second equation from the first
Subtract equation (2) from equation (1) to eliminate $ \phi $:
$ 4.8 - 1.6 = \left(\frac{hc}{\lambda} - \phi\right) - \left(\frac{hc}{2\lambda} - \phi\right)$
$ 3.2 = \frac{hc}{\lambda} - \frac{hc}{2\lambda}$
Factor out $ \frac{hc}{\lambda} $ on the right-hand side:
$ 3.2 = \frac{hc}{\lambda} \left(1 - \frac{1}{2}\right) = \frac{hc}{\lambda} \times \frac{1}{2}$
$ 3.2 = \frac{hc}{2\lambda}$
Thus,
$ \frac{hc}{\lambda} = 6.4 \quad \dots (3)$
Step 4: Find the work function $ \phi $
Use equation (2):
$ 1.6 = \frac{hc}{2\lambda} - \phi$
From the above step, we have $ \frac{hc}{2\lambda} = 3.2 $. Substitute that in:
$ 1.6 = 3.2 - \phi$
So, the work function $ \phi $ is:
$ \phi = 3.2 - 1.6 = 1.6 \text{ eV (or volts in this formulation)}$
Step 5: Relate the threshold wavelength to the work function
The threshold wavelength $ \lambda_{\text{th}} $ is the wavelength for which the photon energy just equals the work function:
$ \frac{hc}{\lambda_{\text{th}}} = \phi$
Substitute $ \phi = 1.6 $:
$ \frac{hc}{\lambda_{\text{th}}} = 1.6$
Or,
$ \lambda_{\text{th}} = \frac{hc}{1.6} \quad \dots (4)$
Compare that with equation (3), which gave $ \frac{hc}{\lambda} = 6.4 $ or $ \lambda = \frac{hc}{6.4} $. Notice that:
$ \lambda_{\text{th}} = \left(\frac{hc}{6.4}\right) \times 4 = 4 \left(\frac{hc}{6.4}\right) = 4\lambda$
Hence, the threshold wavelength of the metal is $ 4\lambda $.
Final Answer
The threshold wavelength of the metal is 4$ \lambda $.
