Question

What wavelength photon would be required to ionize a hydrogen atom in the ground state and give the electron a kinetic energy of 10.7 eV?

Answer 1

Answer:

∝= 51.08 nm

Explanation:

It is necessary that the photon  have the enough energy to get out of the ground state and the energy required to jump out of the ground state = 13.6 eV

so we have the total energy needed:

E= 13.6+10.7

E= 24.3 eV

Planck constant = 6.62*10^-34

with the E formula we could clear the wavelength variable like this....

E= hf

E= h(c/∝)

∝= h*(c/E)

∝=(6.62*10^-34) * [ 3*10^8  /  24.3*(1.6*10^-19)J ]

∝= 5.108*10^-8m

now we just need to convert to a nm distance

∝= 51.08 nm