Question

Calculate the frequency and wavelength of a photon having energy equal to 2 eV.​

Answer 1

Answer:

Answer: The wavelength of a 2 eV photon is given by: l = h c / Eph = 6.625 x 10-34 x 3 x 108/(1.6 x 10-19 x 2) = 621 nm. where the photon energy was multiplied with the electronic charge to convert the energy in Joule rather than electron Volt.

Answer 2

Answer:

Frequency=$4.826 \times 10^{14} \mathrm{~Hz}$

Wavelength=$6216 \AA$

Explanation:

Tip

• The distance between identical spots between two subsequent waves is computed using the wavelength property of a wave.The Greek letter lambda (λ) is used to describe it. As a result, the wavelength is defined as the distance between one wave's crest or trough and the following wave.

As light has the properties of a wave and a particle, it can be expressed in two equations:

$\nu=\lambda f$

$E=h f$

Where,

- $\nu$ is the velocity of the light.

- $\lambda$ is the wavelength of the light.

- $f$is the frequency of the light.

- $E$is the energy of the light wave.

- $h$ is the Planck's Constant i.e.

$6.64 \times 10^{-34} joule \cdot second$

• The number of complete oscillations performed by any wave constituent per unit of time is the frequency of a sinusoidal wave. We can deduce from the definition of frequency that if a body is moving in a periodic pattern, it has completed one cycle after passing through a sequence of events or positions and returning to its original condition.  As a result, frequency is a quantity that describes the rate at which an oscillation or vibration occurs.

Given:

$E=2 \mathrm{eV}=2 \times 1.6 \times 10^{-19}=3.2 \times 10^{-19} \mathrm{~J}$

Find:

Frequency (v)

Wavelength $(\lambda)$

Solution

Using formula (i),

$v=\frac{3.2 \times 10^{-19}}{6.63 \times 10^{-34}}$

$=\frac{3.2 \times 10^{15}}{6.63}$

$={antilog}\left\{\log (3.2)+\log \left(10^{15}\right)-\log (6.63)\right\}$

$=\text { antilog }\{0.5051+15-0.8215\}$

$=\text { antilog }\{14.6836\}$

$=4.826 \times 10^{14} \mathbf{H z}$

Using formula (ii),

$\lambda=\frac{c}{v}=\frac{3 \times 10^{8}}{4.826 \times 10^{14}}$

$=\text { antilog }\{\log 3-\log 4.826\} \times 10^{-6}$

$=\text { antilog }\{0.4771-0.6836\} \times 10^{-6}$

$=\text { antilog }\{\overline{1.7935}\} \times 10^{-6}$

$=0.6216 \times 10^{-6} \mathrm{~m}$

$=6216 \times 10^{-10} \mathrm{~m}$

$=6216 \AA$

Final Answer

i. The frequency of photons is $4.826 \times 10^{14} \mathrm{~Hz}$.

ii. The wavelength of the photon is $6216 \AA$.

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