Pragraph : A hydrogen octum in third excited state makes a transtion to first excited state and emit

photon. This emitted photon is absorbed by He+ ion which was already in seventh excited state. After absorption

of photon He+ ion jumps from seventh excited state to heigher excited state having quantum number ns.

1. The quantum number ns of the state finally populated in He+ ion.

a) 8 b) 10 c) 16 d) 20

2. Now He+ ion jumps to lower state by imitting single photon 'P' of visible light. The energy satatis photon is as possible as close to energy of the obsorbed photon by He+ ion (as mentioned in paragraph). Then wave length of the photon 'P' is nearly.

a) 6760 Angstrom b) 5000A c) 4480A d) 3500 A

3.The ratio of K.E. of the n = 4 electron for H atom to that He⁺ ion is P1 and the ratio of the total energy of n = 6 electron for H atom to that of He⁺ ion is P2 then ratio of P1/P2 is.

a) 1 b) 9/4 c) 4/9 d) 4

Let's break down the problem step by step, focusing on the transitions and energy levels involved for both the hydrogen atom and the He+ ion. This will help us answer the questions regarding quantum numbers, wavelengths, and energy ratios.

Understanding the Transition in Hydrogen

A hydrogen atom in the third excited state (n=4) transitions to the first excited state (n=2). The energy difference between these states corresponds to the energy of the emitted photon. The energy levels of hydrogen can be calculated using the formula:

E_n = -13.6 eV / n²

For n=4:

E_4 = -13.6 eV / 4² = -0.85 eV

For n=2:

E_2 = -13.6 eV / 2² = -3.4 eV

The energy of the emitted photon (ΔE) is:

ΔE = E_2 - E_4 = -3.4 eV - (-0.85 eV) = -2.55 eV

Photon Absorption by He+

The He+ ion is initially in the seventh excited state (n=8). When it absorbs the photon emitted by hydrogen, it transitions to a higher energy state. The energy of the absorbed photon is equal to the energy difference between the two states of the He+ ion. The energy levels for He+ can be calculated similarly:

E_n = -54.4 eV / n²

For n=8:

E_8 = -54.4 eV / 8² = -0.85 eV

To find the new quantum number (ns) after absorption, we need to find the energy level that corresponds to the energy of the absorbed photon:

E_ns = E_8 + ΔE = -0.85 eV + 2.55 eV = 1.7 eV

Now, we find the quantum number that corresponds to this energy:

-54.4 eV / n² = 1.7 eV

n² = 54.4 / 1.7 ≈ 32.12

n ≈ 5.66

Since quantum numbers are whole numbers, we round up to the nearest integer, which gives us n=6. However, since we are looking for the next higher state, we can conclude that the quantum number ns is likely 8, 10, 16, or 20. The closest higher state would be n=10.

Determining the Wavelength of Photon 'P'

Next, we need to find the wavelength of the photon emitted by the He+ ion when it transitions to a lower state. The energy of this photon will be similar to the energy of the absorbed photon, which we calculated as 2.55 eV. The wavelength (λ) can be found using the equation:

λ = hc / E

Where:

• h = Planck's constant (4.1357 x 10^-15 eV·s)
• c = speed of light (3 x 10^8 m/s)

Substituting the values:

λ = (4.1357 x 10^-15 eV·s * 3 x 10^10 cm/s) / 2.55 eV

This calculation will yield a wavelength close to one of the options provided. After performing the calculation, we find:

λ ≈ 4860 Å

However, since the closest option is 5000 Å, we select that as our answer.

Calculating the Ratios of K.E. and Total Energy

For the kinetic energy (K.E.) of an electron in a hydrogen atom and He+ ion, we use the formula:

K.E. = -0.5 * E_n

For n=4 in H:

K.E._H = -0.5 * (-0.85 eV) = 0.425 eV

For n=4 in He+:

K.E._He+ = -0.5 * (-54.4 eV / 16) = 1.7 eV

The ratio P1 of K.E. of n=4 electron for H atom to that of He+ ion is:

P1 = K.E._H / K.E._He+ = 0.425 eV / 1.7 eV = 1/4

For the total energy of n=6 electrons:

E_H = -13.6 eV / 36 = -0.377 eV

E_He+ = -54.4 eV / 36 = -1.51 eV

The ratio P2 of total energy of n=6 electron for H atom to that of He+ ion is:

P2 = E_H / E_He+ = -0.377 eV / -1.51 eV = 1/4

Finally, the ratio P1/P2 is:

P1/P2 = (1/4) / (1/4) = 1

Final Answers

• 1. The quantum number ns of the state finally populated in He+ ion: 10
• 2. The wavelength of the photon 'P': 5000 Å
• 3. The ratio of P1/P2: 1