Question

The kinetic energy of an electron is 5×10^-5 eV. Calculate the wavelength of the wave associated with the electron. The mass of the electron may be taken as 10^-30kg

Answer 1

Explanation:

Kinetic energy of an electron = 5×\sf10^{-5}10

−5

eV

⏭ To Find:

✏ de-Broglie wavelength associated with electron.

⏭ Formula:

✏ Formula of de-Broglie wavelength in terms of kinetic energy and mass of particle is given by

\red{\circ} \: \underline{ \boxed{ \bold{ \sf{ \blue{ \lambda = \dfrac{h}{ \sqrt{2m(KE)}}}}}}} \: \red{ \circ}∘

λ=

2m(KE)

h

∘

⏭ Terms indication:

✏ \lambdaλ denotes wavelength

✏ h denotes plank's constant

✏ m denotes mass of particle

✏ KE denotes kinetic energy associated with particle

⏭ Conversation:

✏ KE = 5×\sf10^{-5}10

−5

eV = 8×\sf10^{-24}10

−24

J

⏭ Calculation:

\begin{lgathered}\mapsto \: \lambda = \dfrac{6.626 \times {10}^{ - 34} }{ \sqrt{2 \times 9.109 \times {10}^{ - 31} \times 8 \times {10}^{ - 24} } } \\ \\ \mapsto \sf \: \lambda = \dfrac{6.626 \times {10}^{ - 34} }{ \sqrt{14.574 \times {10}^{ - 54} } } \\ \\ \mapsto \sf \: \lambda = \frac{6.626 \times {10}^{ - 34} }{3.817 \times {10}^{ - 27} } \\ \\ \mapsto \sf \: \lambda = 1.735 \times {10}^{ - 7} \\ \\ \mapsto \: \underline{ \boxed{ \bold{ \sf{ \orange{ \large{ \lambda = 173.5 \: nm}}}}}} \: \red{ \bigstar}\end{lgathered}

↦λ=

2×9.109×10

−31

×8×10

−24

6.626×10

−34

↦λ=

14.574×10

−54

6.626×10

−34

↦λ=

3.817×10

−27

6.626×10

−34

↦λ=1.735×10

−7

↦

λ=173.5nm

★