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Step-by-Step Solution
Step 1: Write the Photoelectric Equation for Wavelength λ
When the incident radiation has wavelength \lambda , the maximum kinetic energy of the photoelectrons is given by Einstein’s photoelectric equation:
\frac{hc}{\lambda} - \phi = eV
Here,
h is Planck’s constant.
c is the speed of light.
\phi is the work function (in joules).
e is the charge on an electron.
V is the stopping potential (in volts).
Step 2: Write the Photoelectric Equation for Wavelength 3λ
When the incident radiation has wavelength 3\lambda and the stopping potential changes to \frac{V}{4} , the photoelectric equation becomes:
\frac{hc}{3\lambda} - \phi = e \left( \frac{V}{4} \right)
Step 3: Subtract One Equation from the Other
From the two equations, we have:
(1) \frac{hc}{\lambda} - \phi = eV
(2) \frac{hc}{3\lambda} - \phi = \frac{eV}{4}
Subtract equation (2) from equation (1) to eliminate eV and compare the differences on both sides:
\left( \frac{hc}{\lambda} - \phi \right) - \left( \frac{hc}{3\lambda} - \phi \right) = eV - \frac{eV}{4}
\frac{hc}{\lambda} - \frac{hc}{3\lambda} = \phi - \phi + eV - \frac{eV}{4}
\frac{hc}{\lambda} \left(1 - \frac{1}{3}\right) = eV \left(1 - \frac{1}{4}\right)
\frac{hc}{\lambda} \times \frac{2}{3} = eV \times \frac{3}{4}
This relationship can be rearranged or further combined with the original equations as needed. However, an alternative direct approach is to use the given final relationship: from the original solution, we know the difference leads to a simpler form relating \phi directly.
Step 4: Relate Work Function to Incident Energy
Alternatively, from the given solution steps, we use the fact that if we rearrange (1) and (2), we get:
\frac{hc}{3\lambda} - \phi = \frac{1}{4} \left( \frac{hc}{\lambda} - \phi \right)
This simplifies to an expression showing \phi = \frac{hc}{9\lambda} . (The exact steps involve comparing both equations carefully and algebraically.)
Step 5: Determine the Threshold Wavelength \lambda_0
The threshold wavelength \lambda_0 is related to the work function \phi by:
\phi = \frac{hc}{\lambda_0}
We have found \phi = \frac{hc}{9\lambda} . So, equating these:
\frac{hc}{\lambda_0} = \frac{hc}{9\lambda}
Cancel hc on both sides, yielding:
\lambda_0 = 9 \lambda
Step 6: Identify the Value of n
Given that the threshold wavelength \lambda_0 is n \lambda , we compare and see that:
n \lambda = 9 \lambda
Therefore, n = 9 .
Final Answer
The value of n is 9.
