Question

A photon of 19.6 eV energy strikes a H-atom (in its ground state). Find de-Broglie wavelength
of electron ejected from H-atom (in Å).​

Answer 1

Answer is 3.32 $\AA$

Explanation:

De Broglie wavelength ($\lambda$) is given by the equation

$\lambda = \frac {h}{p}$

• h=Planck’s constant whose value is $6.62 \times 10^{-34}$ joule-seconds and

• p = momentum of the particle(here electron)

In terms of kinetic energy(E) momentum(p) can be written as,

$p=(2mE)^{\frac {1}{2}}$

where m=mass of the particle.

Hence $\lambda$ becomes

$\lambda = h(2mE)^{\frac {1}{2}}$ """(1)

Given here,

• E = $13.6 eV = 13.6 \times 1.6 \times 10^{-19}$ joule
• m(mass of electron)= $9.1\times 10^{-31}$ kg

Putting these values in equation (1) we get ,

λ = $0.332 \times 10^{-9}$ meter

= $3.32\times10^{-10}$ meter

= $3.32 \AA$

Answer 2

Answer:

see the attachment .....