Question

The first ionization energy, E , of a helium atom is 3.94 aJ. What is the wavelength of light, in nanometers, that is just sufficient to ionize a helium atom? Values for constants can be found in the Chempendix.

Answer 1

Answer: 50.2 nm

Step-by-step explanation:

The relation between energy and wavelength of light is given by Planck's equation, which is:

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

where,

E = energy of the light = 3.94aJ=
`3.94* 10^(-18)J`
`1aJ=10^(-18)J`

h = Planck's constant =
`6.6* 10^(-34)Js`

c = speed of light =
`3.0* 10^8ms^(-1)`

`\lambda` = wavelength of light = ?

Putting all the values we get:

`3.94* 10^(-18)=(6.6* 10^(-34)* 3.0* 10^8)/(\lambda)`

`\lambda=5.02* 10^(-8)m=50.2nm`
`1nm=10^(-9)m`

Thus the wavelength of light, in nanometers that is just sufficient to ionize a helium atom is 50.2 nm