Question

When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is $6.0 \,V$. This potential drops to $0.6 \,V$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? \([ Take=\frac{h c}{e}=124 \times 10^{-6} Jm C ^{-1}]\)

• A. $1.72 \times 10^{-7} m , 1.20 \, eV$
• B. $1.72 \times 10^{-7} m , 5.60 \,eV$
• C. $3.78 \times 10^{-7} m , 5.60 \,eV$
• D. $3.78 \times 10^{-7} m , 1.20 \, eV$

Answer 1

The Correct Option is A

Let the wavelength of the first light source be $\lambda_1$ and its stopping potential be $V_1 = 6.0 \text{ V}$.

Let the wavelength of the second light source be $\lambda_2$ and its stopping potential be $V_2 = 0.6 \text{ V}$.

We are given that $\lambda_2 = 4\lambda_1$.

Let $\Phi$ be the work function of the metal in eV.

Einstein's photoelectric equation, with energies in eV, is:

$E_{\text{photon}} (\text{eV}) = \Phi (\text{eV}) + V_s$

where $V_s$ is the stopping potential in Volts, and $E_{\text{photon}} (\text{eV}) = \frac{hc/e}{\lambda}$.

For the first source:

$$ \frac{hc/e}{\lambda_1} - \Phi = 6.0 \quad \quad (1) $$

For the second source (using $\lambda_2 = 4\lambda_1$):

$$ \frac{hc/e}{4\lambda_1} - \Phi = 0.6 \quad \quad (2) $$

Let $X = \frac{hc/e}{\lambda_1}$. The equations become:

$$ X - \Phi = 6.0 \quad \quad (1') $$ $$ \frac{X}{4} - \Phi = 0.6 \quad \quad (2') $$

Subtract equation (2') from equation (1'):

$$ \left(X - \Phi\right) - \left(\frac{X}{4} - \Phi\right) = 6.0 - 0.6 $$ $$ X - \frac{X}{4} = 5.4 $$ $$ \frac{3X}{4} = 5.4 $$ $$ X = \frac{5.4 \times 4}{3} = 1.8 \times 4 = 7.2 $$

So, the energy of the incident photons from the first source is $E_1 = X = 7.2 \text{ eV}$.

Substitute $X = 7.2 \text{ eV}$ into equation (1'):

$$ 7.2 - \Phi = 6.0 $$ $$ \Phi = 7.2 - 6.0 $$ $$ \Phi = 1.2 \text{ eV} $$

The work function of the metal is $1.20 \text{ eV}$.

Now, we need to find the wavelength $\lambda_1$. We have $E_1 = \frac{hc/e}{\lambda_1}$, so $\lambda_1 = \frac{hc/e}{E_1}$.

The problem states to take $hc/e = 124 \times 10^{-6} \text{ JmC}^{-1}$. Since $1 \text{ J/C} = 1 \text{ V}$, this means $hc/e = 124 \times 10^{-6} \text{ Vm}$.

Using this value literally:

$$ \lambda_1 = \frac{124 \times 10^{-6} \text{ Vm}}{7.2 \text{ V}} \approx 17.222 \times 10^{-6} \text{ m} = 1.7222 \times 10^{-5} \text{ m} $$

This wavelength does not match the order of magnitude of the options.

However, the standard value for $hc$ is approximately $1240 \text{ eV} \cdot \text{nm}$. Using this (which implies $hc/e \approx 1.240 \times 10^{-6} \text{ Vm}$):

$$ E_1 (\text{eV}) = \frac{hc (\text{eV} \cdot \text{nm})}{\lambda_1 (\text{nm})} $$ $$ 7.2 \text{ eV} = \frac{1240 \text{ eV} \cdot \text{nm}}{\lambda_1 (\text{nm})} $$ $$ \lambda_1 (\text{nm}) = \frac{1240}{7.2} \approx 172.22 \text{ nm} $$ $$ \lambda_1 \approx 172.22 \times 10^{-9} \text{ m} = 1.7222 \times 10^{-7} \text{ m} $$

This wavelength ($1.72 \times 10^{-7} \text{ m}$) and the calculated work function ($1.20 \text{ eV}$) match Option A.

It seems the constant provided in the question ($124 \times 10^{-6} \text{ JmC}^{-1}$) is either a typo or intended to be used in a specific context that is not immediately obvious, as it is 100 times the standard physical value of $hc/e$ (approx $1.24 \times 10^{-6} \text{ JmC}^{-1}$). To match the given options, one must effectively use a standard value for $hc/e$.

Therefore, the wavelength of the first source is $1.72 \times 10^{-7} \text{m}$ and the work function of the metal is $1.20 \text{ eV}$.

The Correct Option is A.