Question

Calculate the energy change that would accompany an electronic transition in a hydrogen atom from n = 2 to n = 3 shell. Also, determine whether radiation is absorbed or emitted during this transition.

Answer 1

`3.03x10^-19~J`

Step-by-step explanation:

We have to use the equation for the hydrogen atom:

Δ

E=
`-2.18 x10^-18~J~((1)/(n^2_f)-(1)/(n^2_i))`

For this case we go from n=2 to n=3, so:
`n^2_f=3` and
`n^2_i=2`.

Now we have to plug the values into the equation:

ΔE=
`-2.18 x10^-18~J~((1)/(3^2)-(1)/(2^2))`

ΔE=
`3.03x10^-19~J`

If the sign is positive we have to give energy to the system. If we have to give energy the system would consume energy. If we have consumption of energy we will have absorption of light.

Answer 2

The energy associated with this transition is 3.04 × 10⁻¹⁹ J.

Step-by-step explanation:

An electron absorbs radiation when it goes from level 2 to level 3. The wavelength associated to this transition can be calculated using Rydberg equation.

`(1)/(\lambda ) =R_(H)((1)/(n_(1)^(2) )-(1)/(n_(2)^(2) )  )`

where,

λ is the wavelength of the radiation

RH is the Rydberg constant for Hydrogen (1.10 × 10⁷ m⁻¹)

n₁ and n₂ are the levels (being n₁ < n₂)

In this case,

`(1)/(\lambda ) =1.10 * 10^(7) m^(-1)  .((1)/(2^(2) )-(1)/(3^(2) )  )\\\lambda = 6.54 * 10^(-7) m`

We can calculate the energy associated to this radiation using Planck-Einstein equation.

E = h . ν = h . c / λ

where,

h is the Planck's constant (6.63 × 10⁻³⁴ J.s)

c is the speed of light (3.00 × 10⁸ m/s)

Then,

`E=h.(c)/(\lambda ) =6.63 * 10^(-34) J.s .(3.00 * 10^(8)m/s)/(6.54 * 10^(-7)m) =3.04 * 10^(-19)J`