Question

the frequency of the spectral line obtained when the electron in n = 3 of hydrogen atom drops to the ground state is​

Answer 1

from Rydberg's equation,

1/λ = R(1/n1² - 1/n2²)

where R is Rydberg's constant.

here, n1 = 1 and n2 = 3

so, 1/λ = R(1/1² - 1/3²)

= R(1 - 1/9) = 8R/9

so, λ = 9/8R

putting value of R = 1.097 × 10^7 m^-1

so, λ = 9/(8 × 1.097 × 10^7)

= 1.0255 × 10^-7 m

now frequency = speed of light/wavelength

= 3 × 10^8/(1.0255 × 10^-7)

= 2.925 × 10^15 Hz

hence, frequency of the spectral line obtained when the electron in n = 3 of hydrogen atom drop to the ground state is 2.925 × 10^15 Hz

Answer 2

Answer:

The Rydberg expression for an electronic transition in the hydrogen atom is:

1λ=R[1n21−1n22]

λ=wavelength

R=1.097×107m−1

n1=3

n2=5

So:

1λ=R(19−125)

1λ=R(0.071)=1.097×107×0.071=7.789×105m−1

The energy of the photon given by this transition is given by:

E=hf

Since c=fλ

E=hcλ

h is the Planck Constant = 6.63×10−34Js

c = the speed of light =3×108m/s

So:

E=6.63×10−34×3×108×7.789

0.071=7.789×105m−1

The energy of the photon given by this transition is given by:

E=hf

Since,

c=fλ

E=hcλ

his the Planck Constant = 6.63×10−34Js

c= the speed of light = 3×108m/s

So:

E=6.63×10-34×3×108×7.789×105J

E=1.531×10−

19J...

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