Question

Question 2.55 Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × 1015 (Hz) [1/32 – 1/n2]

Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Class XI Structure of Atom Page 68

Answer 1

A/C to question,
v = 3.29 × 10^15 ( 1/3² - 1/n²)Hz
we know,
frequency = speed of light/wavelength
given , here
wavelength = 1285 nm = 1.285 × 10^-6 m
speed of light = 3 × 10^8 m/s
frequency ( v ) = 3 × 10^8/1.285 × 10^-6 Hz

3.29 × 10^15 ( 1/3² - 1/n²) Hz = 3 × 10^8/1.285× 10^-6 Hz
( 1/9 - 1/n²) = 3 × 10^8/1.285 × 10^-6 × 3.29 × 10^15
0.1111 - 1/n² = 0.0709
1/n² = 0.1111 - 0.0709 = 0.0402 ≈ 0.04
1/n² = 1/25
1/n² = 1/5²
n = 5

the electrons jumps from n = 5 to n = 3 .e.g., the transition occurs in paschen series and lies infrared region.
Hence, the radiation 1285 nm lies in the infrared region.

Answer 2

Answer:

According to question,

v = 3.29 × 10^15 ( 1/3² - 1/n²)Hz

we know,

frequency = speed of light/wavelength

given , here

wavelength = 1285 nm = 1.285 × 10^-6 m

speed of light = 3 × 10^8 m/s

frequency ( v ) = 3 × 10^8/1.285 × 10^-6 Hz

3.29 × 10^15 ( 1/3² - 1/n²) Hz = 3 × 10^8/1.285× 10^-6 Hz

( 1/9 - 1/n²) = 3 × 10^8/1.285 × 10^-6 × 3.29 × 10^15

0.1111 - 1/n² = 0.0709

1/n² = 0.1111 - 0.0709 = 0.0402 ≈ 0.04

1/n² = 1/25

1/n² = 1/5²

n = 5

the electrons jumps from n = 5 to n = 3 .e.g., the transition occurs in paschen series and lies infrared region.

Hence, the radiation 1285 nm lies in the infrared region.

HOPE IT HELPS YOU....