Question

A photon in a monochromatic beam of light has energy 3 eV. The wavelength is closest to: A) 4960 nm B) 1240 nm C) 413 nm D) 310 nm E) 4.8x10-19 m

Answer 1

`\lambda=4.143* 10^(-7)\ m`

Step-by-step explanation:

It is given that,

Energy of the photon,
`E=3\ eV=3* 1.6* 10^(-19)\ J`

`E=4.8* 10^(-19)\ J`

Let
`\lambda` is the wavelength of the photon. The energy of a photon is given by :

`E=(hc)/(\lambda)`

`\lambda=(hc)/(E)`

`\lambda=(6.63* 10^(-34)* 3* 10^8)/(4.8* 10^(-19))`

`\lambda=4.143* 10^(-7)\ m`

So, the wavelength of the photon is
`\4.143* 10^(-7)\ m`. Hence, this is the required solution.