Answer to Question #280233 in Physics for Anonymous student

Question

A sample of hydrogen atoms is irradiated with light with wavelength 85.5 nm, and electrons are observed leaving the gas.

(a) If each hydrogen atom were initially in its ground level, what would be the maximum kinetic energy in electron volts of these photoelectrons?

(b) A few electrons are detected with energies as much as 10.2 eV greater than the maximum kinetic energy calculated in part (a). Explain. 

Answer 1

a)

$$\frac{hc}{\lambda}=\frac{(6.26\cdot10^{-34})(2.998\cdot10^{8})}{(85.5\cdot10^{-9})}=14.5\ eV$$

$$E_f=-13.6+14.5=0.90\ eV$$

b) n=2

$$0.25(-13.6)=-3.4\ eV$$

greater than the energy of the ground state. 

 If an electron with E= −3.40 eV gains 14.5 eV from the absorbed photon, it will end up with 14.5 eV-3.4 eV =11.1 eV of kinetic energy which is 10.2 eV greater than 0.90 eV.