Question

If ${{\lambda }_{\text{O}}}$ is the threshold wavelength for a photoelectric emission $\lambda$ wavelength of light falling on the surface of the metal, and m mass of the electron. Then the de Broglie wavelength of the emitted electron is:
A. ${{\left( \dfrac{h(\lambda {{\lambda }_{0}})}{2\text{mc(}{{\lambda }_{0}}\text{ - }\lambda )} \right)}^{\dfrac{1}{2}}}$
B. ${{\left( \dfrac{h({{\lambda }_{0}}\text{ - }\lambda )}{2\text{mc(}\lambda {{\lambda }_{0}})} \right)}^{\dfrac{1}{2}}}$
C. ${{\left( \dfrac{h(\lambda \text{ - }{{\lambda }_{0}})}{2\text{mc}\lambda {{\lambda }_{0}})} \right)}^{\dfrac{1}{2}}}$
D. ${{\left( \dfrac{\text{h}\lambda {{\lambda }_{0}}}{2\text{mc}} \right)}^{\dfrac{1}{2}}}$

Answer 1

For this problem, we have to study about the de Broglie wavelength of the electron and the photoelectric equation that is given by Einstein and by deriving the relation of both equations we will get the correct answer.

Complete Step-By-Step Answer:
- In the given question we have to calculate the de Broglie wavelength of the emitted electron.
- As we know that the photoelectric electric effect is a process in which the electron comes out when the light strikes on the metal surface.
- So, according to Einstein, the equation of the phenomenon of the photoelectric effect is:
$\text{E = }{{\text{E}}_{0}}\text{ + }\dfrac{1}{2}\text{m}{{\text{v}}^{2}}$ …..(1)
- Here, ${{\text{E}}_{0}}$ is the energy which has threshold wavelength (The minimum wavelength which is required to release the photon molecules from the metal surface), E is the energy with a wavelength of light.
- Moreover, m is the mass of the electron and v is the velocity.
- Also, the energy of the particle is also equal to the product of Planck's constant and frequency (a division of velocity of light and wavelength).
- So, equation first can be written as:
$\dfrac{\text{hc}}{\lambda }\text{ = }\dfrac{\text{hc}}{{{\lambda }_{0}}}\text{ + }\dfrac{1}{2}\text{m}{{\text{v}}^{2}}$
$\dfrac{1}{2}\text{m}{{\text{v}}^{2}}\text{ = }\dfrac{\text{hc}}{\lambda }\text{ -}\dfrac{\text{hc}}{{{\lambda }_{0}}}\text{ }$
$\dfrac{1}{2}\text{m}{{\text{v}}^{2}}\text{ = hc}\left( \dfrac{1}{\lambda }\text{ -}\dfrac{1}{{{\lambda }_{0}}} \right)$ …. (2)
- Here, we know that the kinetic energy is equal to the $\dfrac{1}{2}\text{m}{{\text{v}}^{2}}$.
- Now, we write the de - Broglie equation that is:
$\lambda \ \text{= }\dfrac{\text{h}}{\text{P}}$, where h is the Planck's constant and P is the momentum.
- The above equation is also given by:
$\lambda \ \text{= }\dfrac{\text{h}}{\sqrt{2\text{ }\times \text{ K}\text{.E}\text{. }\times \text{ m}}}$
- So, from equation (2), we can substitute the value of kinetic energy in the above equation,
$\lambda \ \text{= }\dfrac{\text{h}}{\sqrt{2\text{ }\times \text{ hc}\left( \dfrac{1}{\lambda }\text{ -}\dfrac{1}{{{\lambda }_{0}}} \right)\text{ }\times \text{ m}}}$
$\lambda \ \text{= }\dfrac{\text{h}}{\sqrt{2\text{ }\times \text{ hmc}\left( \dfrac{{{\lambda }_{0}}\text{ - }\lambda }{\lambda {{\lambda }_{0}}} \right)}}$
$\lambda \ \text{= }\sqrt{\dfrac{{{\text{h}}^{2}}}{2\text{ }\times \text{ hmc}\left( \dfrac{{{\lambda }_{0}}\text{ - }\lambda }{\lambda {{\lambda }_{0}}} \right)}}$ = $\lambda \ \text{= }{{\left( \dfrac{\text{h}}{2\text{ }\times \text{ mc}\left( \dfrac{{{\lambda }_{0}}\text{ - }\lambda }{\lambda {{\lambda }_{0}}} \right)} \right)}^{\dfrac{1}{2}}}$
Therefore, option (A) is the correct answer.

Note: The minimum amount of energy which is required to induce the photoelectric effect is known as work function which is denoted by $\phi$. The photons in the above solution are the particles which consist of energy and have zero mass.