First slide

Wave nature of matter

Question

An electron is in an excited state in a hydrogen like atom. It has a total energy of -3.4 eV. The kinetic energy of electron is E and its de Broglie wavelength is $λ$

Moderate

A

$E = 6.8 eV; λ = 6.6 \times 10^{- 10} m$
B

$E = 3.4 eV; λ = 6.6 \times 10^{- 10} m$
C

$E = 3.4 eV; λ = 6.6 \times 10^{- 11} m$
D

$E = 6.8 eV; λ = 6.6 \times 10^{- 11} m$

Solution

$Potential energy = - 2 \times Kinetic energy = - 2 E$
$Total energy = - 2 E + E = - 3.4 eV = - E$
$\begin{array}{rcl} or E & = & 3.4 eV \\ p & = & momentum, m = mass of electron \\ E & = & \frac{p^{2}}{2 m} \\ or p & = & \sqrt{2 mE} \\ = & \sqrt{2 \times 9.1 \times 10^{- 31} \times 3.4 \times 1.6 \times 10^{- 19}} \approx 10^{- 24} \end{array}$
$de Brogile wavelength λ = \frac{h}{p} = \frac{h}{\sqrt{2 m E}} = 6.6 \times 10^{- 10}$

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