Question

Ultraviolet light of wavelength \(\lambda_{1}\) and \(\lambda_{2}\left(\lambda_{2}>\lambda_{1}\right)\) when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energies \(E_{1}\) and \(E_{2}\) respectively. The value of planck's constant can be found from the relation. (1) \(\mathrm{h}=\frac{1}{\mathrm{c}}\left(\lambda_{2}-\lambda_{1}\right)\left(\mathrm{E}_{1}-\mathrm{E}_{2}\right)\) (2) \(\mathrm{h}=\frac{1}{\mathrm{c}}\left(\lambda_{1}+\lambda_{2}\right)\left(\mathrm{E}_{1}+\mathrm{E}_{2}\right)\) (3) \(\mathrm{h}=\frac{\left(\mathrm{E}_{1}-\mathrm{E}_{2}\right) \lambda_{1} \lambda_{2}}{\mathrm{c}\left(\lambda_{2}-\lambda_{1}\right)}\) (4) \(\mathrm{h}=\frac{\left(\mathrm{E}_{1}+\mathrm{E}_{2}\right) \lambda_{1} \lambda_{2}}{\mathrm{c}\left(\lambda_{1}+\lambda_{2}\right)}\) (5) \(\mathrm{h}=\frac{\left(\mathrm{E}_{1}+\mathrm{E}_{2}\right) \lambda_{1} \lambda_{2}}{3 \mathrm{c}\left(\lambda_{2}-\lambda_{1}\right)}\)

Short answer

Option (3): \( h = \frac{(E_1 - E_2) \lambda_1 \lambda_2}{c (\lambda_2 - \lambda_1)} \)

Step-by-step solution

Understand the photoelectric effect

When ultraviolet light with wavelength \(\lambda\) falls on a hydrogen atom in its ground state, it can liberate an electron with a specific kinetic energy \(E\). The energy of the incident photon is related to its wavelength by \(E_\text{photon} = \frac{hc}{\lambda}\).

Write the energy equation for both wavelengths

For the wavelengths \(\lambda_{1}\) and \(\lambda_{2}\) (with \(\lambda_{2} > \lambda_{1}\)), we can write the energy equations as: \[\frac{hc}{\lambda_1} = E_1 + \Phi \quad \text{and} \quad \frac{hc}{\lambda_2} = E_2 + \Phi\] where \(\Phi\) is the work function of the hydrogen atom in the ground state.

Subtract the two equations

Subtract the second equation from the first to eliminate \(\Phi\): \[\frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} = (E_1 + \Phi) - (E_2 + \Phi)\] which simplifies to \[hc \left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right) = E_1 - E_2\]

Simplify the expression

Rewrite the difference of the fractions: \[hc \left(\frac{\lambda_2 - \lambda_1}{\lambda_1 \lambda_2}\right) = E_1 - E_2\]

Solve for Planck's constant

Isolate \(h\) to find its value: \[h = \frac{(E_1 - E_2) \lambda_1 \lambda_2}{c (\lambda_2 - \lambda_1)}\]

Match the answer to the given options

The expression obtained matches option (3). Therefore, \( h = \frac{(E_1 - E_2) \lambda_1 \lambda_2}{c (\lambda_2 - \lambda_1)} \)

Key concepts

Planck's constant

Planck's constant, denoted by the letter \(h\), is a fundamental constant in quantum mechanics. It represents the proportionality factor between the energy of a photon and the frequency of its associated electromagnetic wave. The energy \(E\) of a photon can be calculated using the formula:
\[ E = h u \]
where \( u \) (nu) is the frequency of the electromagnetic wave. This relationship highlights that energy is quantized, meaning it can only take discrete values.
Planck's constant has a value of approximately \(6.626 \times 10^{-34} \text{Js}\). Albert Einstein utilized Planck's constant in his explanation of the photoelectric effect, earning him a Nobel Prize.
Understanding this constant is crucial for grasping various concepts in quantum mechanics and modern physics.

Ultraviolet light

Ultraviolet (UV) light is a type of electromagnetic radiation with wavelengths shorter than visible light, but longer than X-rays. It falls within the wavelength range of about 10 to 400 nanometers.
UV light can be categorized into three types: UVA, UVB, and UVC, each with differing effects and applications.

• **UVA**: Longest wavelength, can cause skin aging and contribute to skin cancer.
• **UVB**: Medium wavelength, responsible for producing vitamin D in the skin and can also cause sunburns.
• **UVC**: Shortest wavelength, highly ionizing and largely absorbed by the Earth's atmosphere.

In the context of the photoelectric effect, UV light provides enough energy to liberate electrons from atoms, making it a vital tool in experiments studying these phenomena.

Kinetic energy

Kinetic energy is the energy an object possesses due to its motion. For an electron being ejected from a material, its kinetic energy \(E_k\) is given by the difference between the energy of the incident photon and the work function \( \Phi \) of the material.
The formula for kinetic energy in this context is:
\[ E_k = h u - \Phi \]
where \(h \u\) is the energy of the incident photon. The higher the energy of the incident photon (which can come from UV light), the greater the kinetic energy of the ejected electron.
Understanding kinetic energy is crucial for interpreting experimental data from photoelectric effect studies and has broader applications in mechanics and other fields of physics.