Question

Carbon dioxide readily absorbs radiation with an energy of 4.67 x 10-20 J. What is the wavelength and frequency of this radiation? Does this radiation fall in the ultraviolet, visible, or infared range?

Answer 1

Answer:

ν = 7.04 × 10¹³ s⁻¹

λ = 426 nm

It falls in the visible range

Explanation:

The relation between the energy of the radiation and its frequency is given by Planck-Einstein equation:

E = h × ν

where,

E is the energy

h is the Planck constant (6.63 × 10⁻³⁴ J.s)

ν is the frequency

Then, we can find frequency,

$\nu = \frac{E}{h}=  \frac{4.67 \times 10^{-20}J  }{6.63 \times 10^{-34}J.s} = 7.04 \times 10^{13} s^{-1}$

Frequency and wavelength are related through the following equation:

c = λ × ν

where,

c is the speed of light (3.00 × 10⁸ m/s)

λ is the wavelength

$\lambda = \frac{c}{\nu } =\frac{3.00 \times 10^{8} m/s }{7.04 \times 10^{13} s^{-1} } =4.26 \times 10^{-6}m.\frac{10^{9}nm }{1m} = 426 nm$

A 426 nm wavelength falls in the visible range (≈380-740 nm)

Answer 2

Frequency of radiation ,ν = 7.04 × 10¹³ s⁻¹

Wavelength of radiation λ = 426 nm

It falls in the visible range.

Equation given by Planck-Einstein:

According to him, the relation between the energy of the radiation and its frequency is given as:

E = h × ν

where,

E is the energy

h is the Planck constant (6.63 × 10⁻³⁴ J.s)

ν is the frequency

1. Calculation of frequency,

$\nu=\frac{E}{h} \\\\\nu=\frac{4.67*10^{-20}J}{6.063*10^{-34}Js} \\\\\nu=7.04*10^{13}s^{-1}$

2. Calculation of wavelength,

c = λ × ν

where,

c is the speed of light (3.00 × 10⁸ m/s)

λ is the wavelength

$c = \lambda * \nu\\\\\lambda =\frac{c}{\nu}\\\\ \lambda=\frac{3.00*10^8m/s}{7.04*10^{13}s^{-1}} =4.26*10^{-6}m\\\\\lambda=426nm$

3. A wavelength of 426 nm  falls in the visible range (380-740 nm)

Find more information about Wavelength here:

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