Question

A photon detector captures a photon with an energy of 4.29 ✕ 10−19 J. What is the wavelength, in nanometers, of the photon?

Answer 1

Answer : The wavelength of photon is,
`4.63* 10^(2)nm`

Explanation : Given,

Energy of photon =
`4.29* 10^(-19)J`

Formula used :

`E=h* \\u`

As,
`\\u=(c)/(\lambda)`

So,
`E=h* (c)/(\lambda)`

where,

`\\u` = frequency of photon

h = Planck's constant =
`6.626* 10^(-34)Js`

`\lambda` = wavelength of photon = ?

c = speed of light =
`3* 10^8m/s`

Now put all the given values in the above formula, we get:

`4.29* 10^(-19)J=(6.626* 10^(-34)Js)* ((3* 10^(8)m/s))/(\lambda)`

`\lambda=4.63* 10^(-7)m=4.63* 10^(-7)* 10^9nm=4.63* 10^(2)nm`

Conversion used :
`1nm=10^(-9)m`

Therefore, the wavelength of photon is,
`4.63* 10^(2)nm`