Question

The first ionization energy, ????, of a hydrogen atom is 2.18 aJ. What is the wavelength of light, in nanometers, that is just sufficient to ionize a hydrogen atom? V

Answer 1

Wavelength of light is 91.2 nm.

Step-by-step explanation:

The first ionization energy of a hydrogen atom is,

`E=2.18aJ\\E=2.18*10^(-18)J`

And Plank's constant,
`h=6.626*10^(-34)Js`

And the speed is,
`c=3*10^(8)m/s`

Now the formula for energy will be,

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

Put all the variable after rearranging for lambda.

`\lambda=(6.626*10^(-34)Js(3*10^(8)m/s))/(2.18*10^(-18)J) \\\lambda=9.12*10^(-8)m\\\lambda=91.2*10^(-9)m\\\lambda=91.2nm`

Therefore the wavelength of light is 91.2 nm.