use the Balmer's equation to determine : (a) The frequency in per second of the radiation corresponding to n=5.
10) The wavelength in nanometers of the line in the Balmer series corresponding to n=7. (c) The value of n
corresponding to the Balmer series line at 380 nm​

Respuesta :

The frequency in per second of the radiation corresponding to n=5 is 6.9 * 10^14 s-1.  The wavelength in nanometers of the line in the Balmer series corresponding to n=7 is  396 nm. The value of n corresponding to the Balmer series line at 380 nm​ is 10.

a) From the Rydberg formula;

1/λ = 1.097 * 10^7 m-1 (1/nf^2 - 1/ni^2)

Since this is the Balmer series, nf =2

1/λ = 1.097 * 10^7 m-1 (1/2^2 - 1/5^2)

1/λ = 1.097 * 10^7 m-1  (0.25 - 0.04)

1/λ = 1.097 * 10^7 m-1 (0.21)

λ = 4.34 * 10^-7 m

But

c = λf

c = speed of light, 3 * 10^8 m/s

f = 3 * 10^8 m/s/ 4.34 * 10^-7

f = 6.9 * 10^14 s-1

b) Again;

1/λ = 1.097 * 10^7 m-1 (1/nf^2 - 1/ni^2)

Since this is the Balmer series, nf =2

1/λ = 1.097 * 10^7 m-1 (1/2^2 - 1/7^2)

1/λ = 1.097 * 10^7 m-1 (0.25 - 0.02)

1/λ = 1.097 * 10^7 m-1 (0.23)

λ = 3.96 * 10^-7 m or 396 nm.

c) Again;

1/λ = 1.097 * 10^7 m-1 (1/nf^2 - 1/ni^2)

Since this is the Balmer series, nf =2

1/380 * 10^-9 = 1.097 * 10^7  (1/2^2 - 1/ni^2)

2631578.947 = 1.097 * 10^7 (1/2^2 - 1/ni^2)

2631578.947 /1.097 * 10^7  = (1/2^2 - 1/ni^2)

0.2399 = 0.25 - 1/ni^2

1/ni^2 = 0.25  - 0.2399

1/ni^2 = 0.0101

ni^2 = 99.0099

ni = 10

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