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Calculate the wavelength in nanometers of the light emitted by a hydrogen atom when it's electron falls from the n= 7 to the n= 4 principal energy level. Recall that the engergy levels of the H atom are given by En = -2.18 x 10 to the negative 18 J ( 1/n to the second power) . ( c= 3.00 x 10 to to the 8 th power m/s ; h=6.63 x 10 to the negative 34 j.s

Respuesta :

Answer:

2.165 x 10^3 nm.

Explanation:

Using Rybergs equation,

1/lambda = R * (1/n1^2 - 1/n2^2)

Where,

R = rybergs constant = 109737.32 cm^-1

n1 = 7

n2 = 4

= 109737.32 * (1/7^2 - 1/4^2)

= 4619.05

Lambda = 2.165 x 10^-4 cm

Since 100 cm = 1m, 1 nm = 10^-9 m

= 2.165 x 10^-4 cm * 1 m/100 cm * 1 nm/10^-9 m

= 2.165 x 10^3 nm.

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