What are the frequency and waveĺength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom ?

Question

Answer

Λ1=R[n121−n221]

λ1=109677.6 cm−1×[221−521]

λ1=23032.3 cm−1

λ=4.34×10−5 cm

Convert unit from cm to nm

1 cm =107 nm

λ=4.34×10−5 cm×1 cm107 nm

λ=434 nm

Answer

Answer

Frequency and Wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom

When an electron in a hydrogen atom transitions from a higher energy level to a lower energy level, it emits a photon of light. The energy of the photon is equal to the energy difference between the two energy levels of the electron. In this case, we are considering the transition from the n = 5 state to the n = 2 state in the hydrogen atom.

Step 1: Calculate the energy difference
To calculate the energy difference, we can use the formula:

ΔE = E_final - E_initial

where ΔE is the energy difference, E_final is the energy of the final state, and E_initial is the energy of the initial state.

The energy of an electron in a hydrogen atom can be calculated using the formula:

E = -13.6 eV / n^2

where E is the energy, n is the principal quantum number, and 13.6 eV is the ionization energy of hydrogen.

For the n = 5 state:
E_initial = -13.6 eV / (5^2)
E_initial = -13.6 eV / 25

For the n = 2 state:
E_final = -13.6 eV / (2^2)
E_final = -13.6 eV / 4

ΔE = E_final - E_initial
ΔE = (-13.6 eV / 4) - (-13.6 eV / 25)
ΔE = -3.4 eV + 0.544 eV
ΔE = -2.856 eV

Step 2: Convert the energy difference to joules
To convert the energy difference from electron volts (eV) to joules (J), we can use the conversion factor:

1 eV = 1.6 x 10^-19 J

ΔE_joules = ΔE_eV * (1.6 x 10^-19 J/eV)
ΔE_joules = -2.856 eV * (1.6 x 10^-19 J/eV)

Step 3: Calculate the frequency
The energy of a photon can be calculated using the formula:

E_photon = h * f

where E_photon is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J.s), and f is the frequency of the photon.

Since the energy of the photon is equal to the energy difference, we can equate the two equations:

ΔE_joules = h * f

Solving for f:

f = ΔE_joules / h
f = (-2.856 eV * (1.6 x 10^-19 J/eV)) / (6.626 x 10^-34 J.s)

Step 4: Calculate the wavelength
The speed of light (c) can be related to the frequency and wavelength of a photon using the formula:

c = λ * f

where c is the speed of light (3 x 10^8 m/s), λ is the wavelength of the photon, and f is the frequency of the photon.

Solving

Similar Class 11 Doubts

What are the frequency and wavelength of a photos emitted during a transition from n=5 to n=2 in H atom?

Calculate energy and wavelength of the photon emitted by hydrogen atom when electron make a transition from n=2 to n=1 given that ionisation potential is 13.6?

What are the frequency and wavelength of a photos emitted during a transition from n=5to n=2 in His atom?

Calculate the energy and frequency of the radiation emitted when an electron jumps from n= 3 to n= 2 in a hydrogen atom.?

the frequency of light emitted for the transition n=4 to n=2 of He+ is equal to the transition in H atom corresponding to which?

Top Courses for Class 11View all

Chemistry Class 11

CUET Mock Test: Humanities Subjects 2025

English Class 11

Mathematics (Maths) Class 11

Psychology Practice Tests: CUET Preparation

Top Courses for Class 11

Top Courses for Class 11

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams