Question 289: Dual Nature of Matter and Radiation

Question

Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with energy E . If $E$ is to be increased to twice the original value, the wavelength changes to $\lambda_1$

Accepted Answer

Let us start with the
photoelectric equation
:
$$ K = h
u - \phi $$
where
$K =$ Kinetic energy of ejected electron
$h =$ Planck's constant
$
u =$ Frequency of incident light
$\phi =$ Work function of the metal
Since $
u = \dfrac{c}{\lambda}$, the equation becomes:
$$ K = \dfrac{hc}{\lambda} - \phi $$
Let the original kinetic energy be $E$:
$$ E = \dfrac{hc}{\lambda} - \phi \qquad \cdots (1) $$
Now, $E$ is
doubled
to $2E$ by changing the wavelength to $\lambda_1$:
$$ 2E = \dfrac{hc}{\lambda_1} - \phi \qquad \cdots (2) $$
Step 1: Express $\phi$ in terms of $E$ and $\lambda$
From (1):
$$ \phi = \dfrac{hc}{\lambda} - E $$
Step 2: Substitute value of $\phi$ in (2):
$$ 2E = \dfrac{hc}{\lambda_1} - \left(\dfrac{hc}{\lambda} - Eight) $$
Simplify:
$$ 2E = \dfrac{hc}{\lambda_1} - \dfrac{hc}{\lambda} + E $$
Bring $E$ to the left:
$$ 2E - E = \dfrac{hc}{\lambda_1} - \dfrac{hc}{\lambda} $$
$$ E = \dfrac{hc}{\lambda_1} - \dfrac{hc}{\lambda} $$
Step 3: Rearranging for $\dfrac{1}{\lambda_1}$
Add $\dfrac{hc}{\lambda}$ to both sides:
$$ E + \dfrac{hc}{\lambda} = \dfrac{hc}{\lambda_1} $$
Divide both sides by $hc$:
$$ \frac{E}{hc} + \frac{1}{\lambda} = \frac{1}{\lambda_1} $$
Recall from equation (1):
$$ E = \frac{hc}{\lambda} - \phi \implies \frac{E}{hc} = \frac{1}{\lambda} - \frac{\phi}{hc} $$
Substitute:
$$ \left(\frac{1}{\lambda} - \frac{\phi}{hc}ight) + \frac{1}{\lambda} = \frac{1}{\lambda_1} $$
$$ \frac{2}{\lambda} - \frac{\phi}{hc} = \frac{1}{\lambda_1} $$
But we are just comparing, observe from earlier:
$$ \frac{E}{hc} + \frac{1}{\lambda} = \frac{1}{\lambda_1} $$
From (1) again, $E = \dfrac{hc}{\lambda} - \phi$:
So,
$$ \frac{E}{hc} = \frac{1}{\lambda} - \frac{\phi}{hc} $$
Plug in:
$$ \frac{1}{\lambda_1} = \bigg( \frac{1}{\lambda} - \frac{\phi}{hc}\bigg) + \frac{1}{\lambda} $$
$$ \Rightarrow \frac{1}{\lambda_1} = \frac{2}{\lambda} - \frac{\phi}{hc} $$
Compare to
$\lambda/2$
:
For $\lambda_1 = \lambda/2$:
$$ \frac{1}{\lambda_1} = \frac{2}{\lambda} $$
But in our equation:
$$ \frac{1}{\lambda_1} = \frac{2}{\lambda} - \frac{\phi}{hc}  \frac{\lambda}{2} $$
Final Answer:
Option C:
$\boxed{\lambda_1 > \lambda/2}$

Answer

$\lambda_1<\lambda / 2$

Answer

$\lambda_1=\lambda$

Answer

$\lambda_1>\lambda / 2$

Answer

$\lambda_1=\lambda / 2$

Dual Nature of Matter and Radiation

Light of wavelength \lambda strikes a photoelectric surface and electrons are ejected with energy E . If E is to be increased to twice the original value, the wavelength changes to \lambda_1

Choose the correct answer: