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A hydrogen atom in an n = 2 state absorbs a photon,

  1. What should be the photon wavelength to cause the electron to jump to an n = 4 state?
  2. What wavelength photons might be emitted by the atom following this absorption?

Short Answer

Expert verified

(a) Wavelength of the photon to cause the jump, will be 486 nm .

(b)The possible wavelengths emitted are

=486nm,103nm,656nm,122nm,1875nm,97.2nm.

Step by step solution

01

A concept:

Energy levels or energy state is any discrete value from a set of total energy values for a force-limited subatomic particle in a limited space, or for a system of such particles, such as an atom or a nucleus.

02

(a) Energy difference between state n = 4  and n = 1 :

The energy of a state is defined by the principal quantum number and the difference in energy between two states is said to be the energy difference between two states.

As you know that energy of a state is

En=-13.6eVn2n=1,2,3,.......

The energy difference between orbit and orbit 1 is given by,

role="math" localid="1659326526082" E=TheEnergyorbit4-TheEnergyoforbit1\=E4-E1=-13.6eV142-122E=2.55eV

鈥.. (1)

03

(a) Wavelength of photon to make electron jump  from n = 1 to n = 4 :

For calculating the wavelength, you will use the equation given below

E=hc 鈥.. (2)

Here, is the wavelength, E is the energy of photon, h is the Planck鈥檚 constant, and c is the speed of light.

The Plank鈥檚 constant, hc = 1240 eV.nm

Substitute 2.55eV for E and 1240 eV. nm for hc into equation (2).

role="math" localid="1659326706949" 1240eV.nm=2.55eV=486nm

04

(b) Possible paths of the electron after this absorption:

The electron can jump from n = 4 to n = 1 in the following different paths:

1. 42then21

2. 43then32then21

3. 41

4.43then31

Hence, the emitted wavelengths will be corresponding to the above-mentioned jumps.

05

Possible emitted wavelengths:

From step 3, the unique combination of jumps is,

42,21,43,32,41and31.

Hence, wavelength corresponding to only these 6 jumps can be emitted.

As known that,

1=1.097107m-11nf2-1ni2

Where, is the Wavelength emitted, nf is the final orbit, and niis the initial orbit.

Now, for42:

1=1.097107m-11(2)2-142=1.097107m-10.25-0.0625=0.02056875107m-1

=10.02056875107m-1=4.86107m=4.8610-9m=486nm

For 43:

1=1.097107m-11(3)2-142=1.097107m-10.111-0.0625=0.0532045107m-1

=10.0532107m-1=18.75107m=18.7510-9m=1875nm

For 41:

1=1.097107m-11(1)2-142=1.097107m-11-0.0625=1.02843107m-1

=11.02843107m-1=0.972107m=0.97210-9m=97.2nm

For 21:

localid="1659327531207" 1=1.097107m-11(1)2-122=1.097107m-11-0.025=0.82275107m-1=10.82275107m-1=1.22107m=12210-9m=122nm

For 32:

1=1.097107m-11(1)2-132=1.097107m-10.25-0.111=0.1524107m-1=10.1524107m-1=6.56107m=65610-9m=656nm

For 31:

1=1.097107m-1112-132=1.097107m-11-0.111=0.97523107m-1

=10.97523107m-1=1.0310-7m=10310-9m=103nm

Hence, the possible wavelengths emitted are as below.

=486nm,103nm,656nm,122nm,1875nm,97.2nm

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