Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,371 questions

17,804 answers

1,438 comments

36,916 users

help with physics

0 votes

In an hydrogen atom, an electron falls from the n = 4 level to the n = 2 level.

 

image

 

Use the energy level diagram for hydrogen above to help you answer the following questions:

a) During this transition, is a photon emitted or absorbed?

b) What is the change in energy of the electron, in eV's?

c) The wavelength of the emitted or absorbed photon is _____ x 10-7 m. 

 

2 -a)What is the shortest wavelength photon that is emitted in the hydrogen atom?


b)What transition emits this photon?


Hint: If the wavelength is small, then the energy is large. Looking at the energy level diagram will also help you.

asked Dec 11, 2014 in PHYSICS by anonymous
reshown Dec 11, 2014 by yamin_math

5 Answers

0 votes

1a)

During this transition, photon jumps from the higher energy band (n=4) to the lower energy band (n=2).

Energy is emitted when the electron moves to an orbit of lower energy state.

So photon emitted energy.

answered Dec 11, 2014 by Shalom Scholar
0 votes

1b)

Initially electron at n = 4 level.

So initial energy Ei = - 0.85

At the end electron is at n = 2 level.

So final energy Ef = - 3.4

The change in energy of the electron E = Ef - Ei

E = 3.4 – ( 0.85 ) = 3.4 + 0.85 = 2.55 eV.

The change in energy of the electron is  2.55 eV.

answered Dec 11, 2014 by Shalom Scholar
0 votes

1c)

E = 2.55 eV  ( From problem 1b )

1 eV = 1.60217657 × 10-19 J

E = 2.55 eV = 2.55×1.60217657 × 10-19 = 4.08555 × 10-19  J

Formula : E = hc/ λ

λ = hc/E

Where E = energy.

          c = Velocity of light = 3 × 108 m/s

          h = Plank’s constant = 6.63 × 10-34 J.s

Substitute :  c = 3 × 108  , h =  6.63 × 10-34 and E = 4.08555 × 10-19

λ = (6.63 × 10-34)( 3 × 108)/( 4.08555 × 10-19  )

λ = (6.63 × 3/4.08555)(10-34+8+19)

λ = 4.8684 × 10-7 m.

The wavelength of the emitted photon is 4.8684 × 10-7 m.

answered Dec 11, 2014 by Shalom Scholar
0 votes

2a)

There are only four state transitions that will result in a Balmer photon being emitted :

Those states are

1) n = ∞  →   n = 2

2) n = 5   →   n = 2

3) n = 4   →   n = 2

4) n = 3   →   n = 2

The wavelength of the photons is given by

image

image

image

R is constant. So R value will not impact on the value of λ.

The shortest λ will be given for the largest value of (1/n1² - 1/n2²).

If n1 is small then (1/n1²) is large. So n1 must be as small as possible.

n1 = 2   ( by considering  Balmer series )

If n2 is high then (1/n2²) is small. So n2 must be as large as possible.

n2 = ∞

image

Infinity value of n2 corresponds to the ionized state of the hydrogen atom.

The shortest wavelength photon that is emitted in the hydrogen atom is 4/R.

answered Dec 11, 2014 by Shalom Scholar
0 votes

2b)

The shortest wavelength photon that is emitted in the hydrogen atom is at

the transition from the energy band n = ∞  to the  energy band  n = 2.

answered Dec 11, 2014 by Shalom Scholar

Related questions

asked Dec 14, 2014 in PHYSICS by anonymous
asked Jun 28, 2017 in PHYSICS by anonymous
asked Jan 26, 2016 in PHYSICS by anonymous
asked Jan 26, 2016 in PHYSICS by anonymous
asked Jan 26, 2016 in PHYSICS by anonymous
asked Jan 26, 2016 in PHYSICS by anonymous
asked Jan 25, 2016 in PHYSICS by anonymous
asked Nov 29, 2015 in PHYSICS by anonymous
...