Answer:
a) λ = 121.13 nm (Ultraviolet)
b) λ = 91.966 nm (Ultraviolet)
c) λ = 486 nm (Visible)
d) λ = 433.91 nm (Visible)
Step-by-step explanation:
a)
Electromagnetic radiations emitted from hydrogen atom only when an electron makes transition from outer to inner shell. Electromagnetic radiations has different region depending upon the transition of electron.
When Electron moves from second shell to first shell then spectral lines is in the ultraviolet region which belongs to Lyman series.
The given principal quantum number is
n = 2 → n = 1
First calculate the difference between energy levels of initial and final from the formula given below
=ΔE= -2.18 × 10⁻¹⁸ (
)
ΔE = -2.18 × 10⁻¹⁸
ΔE = -2.18 × 10⁻¹⁸ × 0.75
ΔE = -1.64 × 10⁻¹⁸ J
After calculating the change in energy, use formula for energy of the photon given below
ΔE =
where
h = Plank's constant = 6.626 × 10⁻³⁴Js
c = speed of light = 2.998 × 10⁸ m/s
λ = Wavelength = ?
λ =
λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) / (1.64 × 10⁻¹⁸ )
λ = 1.2113 × 10⁻⁷ m
λ = 121.13 nm
Ultraviolet spectrum has range of wavelength from 91 nm to 122 nm and 91 nm < 121.13 nm < 122 nm. So, It belongs to Ultraviolet region.
b)
When Electron moves from thirty-second shell to first shell then spectral lines is in the ultraviolet region which belongs to Lyman series.
The given principal quantum number is
n = 32 → n = 1
First calculate the difference between energy levels of initial and final from the formula given below
=ΔE= -2.18 × 10⁻¹⁸ (
)
ΔE = -2.18 × 10⁻¹⁸
ΔE = -2.18 × 10⁻¹⁸ × 0.99
ΔE = -2.16× 10⁻¹⁸ J
After calculating the change in energy, use formula for energy of the photon given below
ΔE =
where
h = Plank's constant = 6.626 × 10⁻³⁴Js
c = speed of light = 2.998 × 10⁸ m/s
λ = Wavelength = ?
λ =
λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) / (2.16 × 10⁻¹⁸ )
λ = 9.1966 × 10⁻⁸ m
λ = 91.966 nm
Ultraviolet spectrum has range of wavelength from 91 nm to 122 nm and 91 nm < 91.966 nm < 122 nm. So, It belongs to Ultraviolet region.
c)
When Electron moves from fourth shell to second shell then spectral lines is in the visible region which belongs to Balmer series.
The given principal quantum number is
n = 4 → n = 2
First calculate the difference between energy levels of initial and final from the formula given below
=ΔE= -2.18 × 10⁻¹⁸ (
)
ΔE = -2.18 × 10⁻¹⁸
ΔE = -2.18 × 10⁻¹⁸ × 0.1875
ΔE = -0.40875 × 10⁻¹⁸ J
After calculating the change in energy, use formula for energy of the photon given below
ΔE =
where
h = Plank's constant = 6.626 × 10⁻³⁴Js
c = speed of light = 2.998 × 10⁸ m/s
λ = Wavelength = ?
λ =
λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) / (0.40875 × 10⁻¹⁸ )
λ = 4.8599 × 10⁻⁷ m
λ = 486 nm
Visible spectrum has range of wavelength from 365 nm to 656 nm and 365 nm < 486 nm < 656 nm. So, It belongs to visible region.
d)
When Electron moves from fifth shell to second shell then spectral lines is in the visible region which belongs to Balmer series.
The given principal quantum number is
n = 5 → n = 2
First calculate the difference between energy levels of initial and final from the formula given below
=ΔE= -2.18 × 10⁻¹⁸ (
)
ΔE = -2.18 × 10⁻¹⁸
ΔE = -2.18 × 10⁻¹⁸ × 0.21
ΔE = -0.4578 × 10⁻¹⁸ J
After calculating the change in energy, use formula for energy of the photon given below
ΔE =
where
h = Plank's constant = 6.626 × 10⁻³⁴Js
c = speed of light = 2.998 × 10⁸ m/s
λ = Wavelength = ?
λ =
λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) / (0.4578 × 10⁻¹⁸ )
λ = 4.3391 × 10⁻⁷ m
λ = 433.91 nm
Visible spectrum has range of wavelength from 365 nm to 656 nm and 365 nm < 433.91 nm < 656 nm. So, It belongs to visible region.