Question

Waveleng
th of photon which have energy

equal to average of energy of photons with λ1

= 4000Å and λ2 = 6000Å will be

1) 5000 Å 2) 4800 Å

3) 9600 Å 4) 2400 Å​

Answer 1

$\red\bigstar$ Explanation $\red\bigstar$

$\leadsto$ Given:-

The energy of a photon is the average of photons have wavelengths $\{lambda}_{1}$ and $\{lambda}_{2}$

$\leadsto$ To Find:-

Wavelength of that photon

$\leadsto$ Solution:-

We know that E = $\dfrac{hc}{[tex]\lambda$}, where h is Planck's constant which is equal to 6.626 $\times$ ${10}^{-34}$ J/Hz

and c is speed of light which is equal to 3 $\times$ ${10}^{8}$ m/s and $\lambda$ is the wavelength of photon

We also know that $\lambda$ (Wavelength) and $\nu$ (Frequency) are inversely proportional,

Therefore,

E = h $\times$ $\nu$

First let's find the energy of the photon which $\lambda$ = 4000 Å = 4000 $\times$ $\{10}^{-10}$

E1 = $\frac{6.626 [tex]\times$ $\{10}^{-34}$ $\times$ 3 $\times$ $\{10}^{8}$}{4000 $\times$ $\{10}^{-10}$}

E1 = 4.97 $\times$ ${10}^{-19}$ J

First let's find the energy of the photon which $\lambda$ = 6000 Å = 6000 $\times$ $\{10}^{-10}$

E2 = $\frac{6.626 [tex]\times$ $\{10}^{-34}$ $\times$ 3 $\times$ $\{10}^{8}$}{6000 $\times$ $\{10}^{-10}$}

E2 = 6.63 $\times$ ${10}^{-19}$ J

Now let's find the Average of E1 and E2 which is the energy of the photon

E = (E1 + E2)/2

E = (4.97 $\times$ ${10}^{-19}$ + 6.63 $\times$ ${10}^{-19}$)/2

E = 5.8 $\times$ ${10}^{-19}$ J

Wavelength of the photon = hc/E = $\frac{6.626 [tex]\times$ ${10}^{-34}$ $\times$ 3 $\times$ ${10}^{8}$}{5.8 $\times$ ${10}^{-19}$}

Wavelength of the photon = 3427 Å