Bohr S Model For Hydrogen Atom

Step 1: Recall the Formula for Energy Levels in Hydrogen-like Atoms
The energy of an electron in a hydrogen-like atom is given by:

where is the atomic number and is the principal quantum number.

Step 2: Calculate the Energy Difference
The energy released ( ) when an electron jumps from an initial state ( ) to a final state ( ) is given by the difference in energy levels:

In this case, , , and . Substituting these values, we get:

Conclusion: The energy released in the process is 40.8 eV (Option 1).

Step 1: Apply Einstein’s Photoelectric Equation

Einstein’s photoelectric equation states:

where is the maximum kinetic energy of the emitted photoelectrons, is Planck’s constant, is the frequency of the incident light, and is the threshold frequency.

Step 2: Relate Kinetic Energy and Velocity

The maximum kinetic energy is also given by:

where is the mass of the electron and is its velocity.

Step 3: Calculate

When :

Step 4: Calculate

When :

Step 5: Find the Ratio

Dividing the equation for by the equation for , we get:

Conclusion: The ratio of to is (Option 4).

The problem states that in a Franck-Hertz experiment with hydrogen, the first dip in the current-voltage graph occurs at 10.2 V. We need to calculate the wavelength of the light emitted when a hydrogen atom, excited by this energy, returns to its ground state.

Concept Used:

The solution is based on the principles demonstrated by the Franck-Hertz experiment and the Bohr model of the atom.

1. Franck-Hertz Experiment: The dips in the current-voltage graph correspond to the voltages at which the electrons have just enough kinetic energy to cause an inelastic collision with the gas atoms. This energy is transferred to the atom, exciting it from its ground state to a higher energy level. The first dip corresponds to the first excitation energy of the atom.
2. Excitation Energy: The energy ( ) absorbed by an atom from an electron accelerated through a potential is given by , where is the charge of an electron. This energy corresponds to the difference between the ground state energy ( ) and the first excited state energy ( ).
3. Photon Emission: When the excited atom de-excites and returns to its ground state, it emits a photon. The energy of this photon ( ) is equal to the energy it absorbed during excitation, .
4. Planck's Relation: The energy of a photon is related to its wavelength ( ) by the equation: where is Planck's constant and is the speed of light.

Step-by-Step Solution:

Step 1: Determine the first excitation energy of the hydrogen atom.

The first dip in the current is observed at an accelerating voltage of . This voltage is the first excitation potential. The energy absorbed by a hydrogen atom during the inelastic collision is:

This is the energy required to excite a hydrogen atom from its ground state ( ) to its first excited state ( ).

Step 2: Determine the energy of the emitted photon.

When the excited hydrogen atom de-excites from the first excited state back to the ground state, it releases this energy in the form of a single photon. Therefore, the energy of the emitted photon is equal to the excitation energy.

Step 3: Calculate the wavelength of the emitted photon.

Using Planck's relation, we can find the wavelength corresponding to the photon's energy:

We are given the value of the product .

Step 4: Substitute the values and compute the result.

Rounding the result to the nearest integer, as is common for such problems, we get:

Thus, the wavelength of light emitted by the hydrogen atom when excited to the first excitation level is 122 nm.

To solve this question, let's analyze each statement separately:

Statement I: "The dimensions of Planck’s constant and angular momentum are same."

Planck's constant has the dimensional formula . Angular momentum (of a particle with mass , revolving with velocity , at a radius ) also has the dimensional formula given by , leading to .

Thus, Planck’s constant and angular momentum indeed have the same dimensions. So, Statement I is correct.

Statement II: "In Bohr’s model, electron revolves around the nucleus in those orbits for which angular momentum is an integral multiple of Planck’s constant."

According to Bohr’s model, the angular momentum of an electron is quantized and given by:

where is a positive integer.

This expression shows that angular momentum is an integral multiple of (also known as reduced Planck's constant ), not Planck's constant itself. Hence, Statement II is incorrect because it states that angular momentum is an integral multiple of without the factor.

Given the analysis:

• Statement I is correct.
• Statement II is incorrect.

Therefore, the correct answer is: Statement I is correct but Statement II is incorrect.