Bohr S Model For Hydrogen Atom

Energy of One Photon
The energy of a single photon is given by:

where and . Substitute the values:

Step 2: Energy of One Mole of Photons
The energy of one mole of photons is:

where . Substitute the values:

Approximate to the nearest integer:

Conclusion: The \textbf{energy} of one mole of photons is .

The correct answer is 2.

The energy of an electron in the -th Bohr orbit is given by the formula: For the first orbit ( ): For the third orbit ( ): Hence, the energy in the third orbit is of the energy in the first orbit.

Solution:
Statement I: Boron is extremely hard indicating its high lattice energy.
Boron forms a giant covalent structure rather than an ionic lattice. Its hardness is due to its strong covalent network, not high lattice energy. Therefore, Statement I is incorrect.

Statement II: Boron has the highest melting and boiling point compared to its other group members.
Boron has a very high melting and boiling point due to its strong covalent network structure. This is true compared to the other group 13 elements. Therefore, Statement II is correct.

Therefore, Statement I is incorrect but Statement II is correct.

Correct Answer: (1) Statement I is incorrect but Statement II is correct

Given:

• Activation energy for hydrogen adsorption on platinum ( ): 30 kJ/mol =
• Activation energy for hydrogen adsorption on nickel ( ): 41.4 kJ/mol =
• Temperature ( ): 300 K
• Gas constant ( ): 8.3 J/K/mol

Formula:

The rate of chemisorption is proportional to . Therefore, the ratio of rates is:

Step-by-Step Calculation:

1. Difference in activation energies:
2. Substitute into the exponent:
3. Calculate:
4. Ratio of rates (logarithmic form):
5. Convert to base-10 logarithm:

Conclusion:

The logarithm of the ratio of chemisorption rates is approximately 2.

The problem states that for a hydrogen atom, the total energy of an electron in the first excited state is . We need to find the kinetic energy (K.E.) of this electron and express it in the form of .

Concept Used:

According to the Bohr model for the hydrogen atom, the total energy (E) of an electron in a specific orbit is the sum of its kinetic energy (K.E.) and potential energy (P.E.). The relationships between these quantities are given by:

From these, a direct relationship between the total energy and the kinetic energy can be derived:

This means that the kinetic energy of the electron is the negative of its total energy. Since kinetic energy must be a positive value, the total energy of a bound electron is always negative.

Step-by-Step Solution:

Step 1: Identify the given total energy of the electron.

The total energy of the electron in the first excited state is given as:

Step 2: Apply the formula relating kinetic energy and total energy.

The formula is:

Step 3: Substitute the given value of E into the formula to calculate the kinetic energy.

Final Computation & Result:

Step 4: Express the calculated kinetic energy in the required format.

The problem asks for the value of where the kinetic energy is .

Solving for :

The value of is an integer, so no rounding is necessary.

The value of is 34.

To determine the de-Broglie wavelength of the electron in the second orbit of the hydrogen atom, we start by analyzing the quantization conditions for electrons in Bohr's model and the de-Broglie wavelength relation.

The de-Broglie wavelength is given by:

where:

• is Planck's constant.
• is the mass of the electron.
• is the velocity of the electron.

For an electron in the nth orbit of a hydrogen atom, the quantized angular momentum condition is:

Solving for , we get:

Substituting in the expression for the radius of the th orbit:

where is the Bohr radius, and substituting this into the expression for :

Plug this velocity back into the de-Broglie wavelength equation:

which simplifies to:

The terms simplify further to:

In the second orbit, where , we substitute :

This matches with the option: