Question

Calculate the wavelength of the light emitted when an electron in a hydrogen atom makes each of the following transitions.
a. n = 2 --> n = 1 ?
b. n = 3 --> n = 1 ?
c. n = 4 --> n = 2 ?
d. n = 5 --> n = 2 ?

Answer 1

Answer:

a. 103 nm

b. 122 nm

c. 486 nm

d 434  nm

Explanation:

We can calculate the wavelength emitted in the hydrogen atom electron transitions from Rydberg´s equation:

1/λ = Rh x ( 1/n₁² - 1/n₂² ) where   λ is the wavelength        

                                                    Rh is Rydberg´s constant 1.097 x 10⁷/m

                                                    n₁, n₂ the principal quantum numbers in                  

                                                    the transitions with n₁ < n₂  by convention.

We have all the data required to solve this question. The result will be in meters, being so small it will be converted to nanometers as is customary in this kind of problems.

a. n₁ = 1 , n₂ = 2

1/λ = 1.097 x 10⁷/m  x ( 1/1² - 1/2² ) = 1.097 x 10⁷/m ( 0.75 ) = 8.228 x 10⁶ / m

Taking inverse to both sides of the equation:

λ = 1 /8.228 x 10⁶ /m = 1215 x 10⁻⁷ m x  1 x 10⁹ nm/ m = 122 nm

b. n₁ = 1 , n₂ = 3

1/λ = 1.097 x 10⁷/m  x ( 1/1² - 1/3² ) = 1.097 x 10⁷/m ( 0.89 ) = 9.751 x  10⁶ / m                                                

λ = 1 /9.751 x 10⁶ /m = 1.026 x 10⁻⁷ m  x 1 x 10⁹ nm/ m = 103 nm

c.  n₁ = 2 , n₂ = 4

1/λ = 1.097 x 10⁷/m  x ( 1/2² - 1/4² ) = 1.097 x 10⁷/m ( 0.19 ) = 2.057 x 10⁶ / m

λ = 1 /2.057 x 10⁶ /m =4.862 x 10⁻⁷ m  x 1 x 10⁹ nm/ m = 486 nm

d.  n₁ = 2 , n₂ = 5

1/λ = 1.097 x 10⁷/m  x ( 1/2² - 1/5² ) = 1.097 x 10⁷/m ( 0.21 ) = 2.304 x 10⁶ / m

λ = 1 /2.304 x 10⁶ /m =4.341 x 10⁻⁷ m  x 1 x 10⁹ nm/ m = 434 nm

Notice how this answers are what we are expecting:

1. The transions to n= 1 are more energetic than to n = 2 since the shorter the wavelength the more energetic is the radiation.

2. Within the same energy  level, the greater n₂, the more energetic is the energy emitted.

Answer 2

Answer:

The wavelength of light emitted in (a) is 122 nm, (b) is 103 nm, (c) is 486 nm, and for (d) is 434 nm.

Explanation:

The equation for the calculation of light emitted on electron transfer is:

$\rm \frac{1}{\lambda}\;=\;R\left ( \frac{1}{n_1^2}\;-\;\frac{1}{n_2^2} \right )$

where,

$\rm \lambda$ = wavelength of light emitted

R =  Rydberg´s constant = 109677

$\rm n_1$ = principal quantum number of initial shell

$\rm n_2$ = principal quantum number of final shell

Putting the given values in the equation above:

(a) For, n = $\rm 2\;\rightarrow\;1$

$\rm \lambda$ = 122 nm

(b) For, n = $\rm 3\;\rightarrow\;1$

$\rm \lambda$ = 103 nm

(c) For, n = $\rm 4\;\rightarrow\;2$

$\rm \lambda$ = 486 nm

(d) For, n = $\rm 5\;\rightarrow\;2$

$\rm \lambda$ = 434 nm.

The maximum wavelength light is emitted in transition from higher shells.

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