p = momentum
This is de Broglie wave equation.
This wave will be having energy in terms of $\lambda$ as:
Analyze the problem. List out the given given quantities and find out which quantity is missing.
Step 2 :
If momentum of photon is given then use the formula
$\lambda$ = $\frac{h}{p}$
where h = Plancks constant,
p = momentum of photon.
If energy is given, then use formula
$\lambda$ = $\frac{h c}{E}$
Here E = Energy of Photon and
c = Velocity of light.
If Wavelength is given you can any of these formula and can get desired quantities of the two.
Calculate the de Broglie wavelength for an electron having velocity 5 $\times$ 105 m/s.If mass is 9.1 $\times$ 10-31 Kg.
Step 1 :given: Mass m = 9.1 $\times$ 10-31 Kg,
Velocity V = 5 $\times$ 105 m/s,
Momentum of electron p = mv = 9.1 $\times$ 10-31 Kg $\times$ 5 $\times$ 105 m/s.
= 45.5 $\times$ 10-26 kgm/s.
Step 2 :If momentum of photon is given then use the formula
$\lambda$ = $\frac{h}{p}$
= $\frac{6.624 \times 10^{-34}}{45.5 \times 10^{-26}}$
= 1.45 $\times$ 10-9 m.
Answer :$\lambda$ = 14.5 $A^{\circ}$
Calculate the momentum and energy for an electron if de-broglie wavelength is 0.24 $A^{\circ}$
Step 1 :Mass m = 9.1 $\times$ 10-31 Kg,
Wavelength $\lambda$ = 0.24 $\times$ 10-10 m
Step 2 :Using the formula
$\lambda$ = $\frac{h}{p}$
Then momentum is given by
p = $\frac{h}{\lambda}$
= $\frac{6.624 \times 10^{-34}}{0.24 \times 10^{-10}}$
= 2.76 $\times$ 10-26 kgm/s.
Energy is given by
E = $\frac{hc}{\lambda}$
= $\frac{6.624 \times 10^{-34} \times 3 \times 10^{8}}{0.24 \times 10^{-10}}$
= 8.28 $\times$ 10-15 eV.
Answer :Momentum p = 2.76 $\times$ 10-26 kgm/s.
Energy E = 8.28 $\times$ 10-15 ev.