Solveeit Logo
Login

Question

Question: What should be the increase in kinetic energy of an electron in order to decrease its de-Broglie wav...

What should be the increase in kinetic energy of an electron in order to decrease its de-Broglie wavelength from 100 nm to 50 nm?

[h = 6 x 10-34 J.s, mₑ = 9 × 10-31 kg, 1 eV = 1.6 × 10-19]

A

0.451 eV

B

3.75 × 10-4 eV

C

4.51 x 10-3 eV

D

0.0375 eV

Answer

3.75 × 10-4 eV

Explanation

Solution

The de-Broglie wavelength (λ\lambda) of a particle is related to its momentum (pp) by the formula:

λ=hp\lambda = \frac{h}{p}

where hh is Planck's constant.

The kinetic energy (KEKE) of a particle with mass mm and momentum pp is given by:

KE=p22mKE = \frac{p^2}{2m}

From this, we can express momentum as p=2mKEp = \sqrt{2mKE}.

Substitute the expression for pp into the de-Broglie wavelength formula:

λ=h2mKE\lambda = \frac{h}{\sqrt{2mKE}}

To find the kinetic energy in terms of wavelength, rearrange the formula:

λ2=h22mKE\lambda^2 = \frac{h^2}{2mKE}

KE=h22mλ2KE = \frac{h^2}{2m\lambda^2}

Let KE1KE_1 be the initial kinetic energy corresponding to λ1=100 nm\lambda_1 = 100 \text{ nm}, and KE2KE_2 be the final kinetic energy corresponding to λ2=50 nm\lambda_2 = 50 \text{ nm}.

Given values:

h=6×10−34 J.sh = 6 \times 10^{-34} \text{ J.s}

me=9×10−31 kgm_e = 9 \times 10^{-31} \text{ kg}

λ1=100 nm=100×10−9 m=1×10−7 m\lambda_1 = 100 \text{ nm} = 100 \times 10^{-9} \text{ m} = 1 \times 10^{-7} \text{ m}

λ2=50 nm=50×10−9 m=5×10−8 m\lambda_2 = 50 \text{ nm} = 50 \times 10^{-9} \text{ m} = 5 \times 10^{-8} \text{ m}

1 eV=1.6×10−19 J1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}

The increase in kinetic energy is ΔKE=KE2−KE1\Delta KE = KE_2 - KE_1.

ΔKE=h22meλ22−h22meλ12\Delta KE = \frac{h^2}{2m_e\lambda_2^2} - \frac{h^2}{2m_e\lambda_1^2}

ΔKE=h22me(1λ22−1λ12)\Delta KE = \frac{h^2}{2m_e} \left( \frac{1}{\lambda_2^2} - \frac{1}{\lambda_1^2} \right)

Calculate the common factor h22me\frac{h^2}{2m_e}:

h22me=(6×10−34 J.s)22×(9×10−31 kg)=36×10−6818×10−31=2×10−37 J.m2\frac{h^2}{2m_e} = \frac{(6 \times 10^{-34} \text{ J.s})^2}{2 \times (9 \times 10^{-31} \text{ kg})} = \frac{36 \times 10^{-68}}{18 \times 10^{-31}} = 2 \times 10^{-37} \text{ J.m}^2

Calculate the term in the parenthesis:

1λ22=1(5×10−8 m)2=125×10−16 m2=12.5×10−15 m2=0.4×1015 m−2=4×1014 m−2\frac{1}{\lambda_2^2} = \frac{1}{(5 \times 10^{-8} \text{ m})^2} = \frac{1}{25 \times 10^{-16} \text{ m}^2} = \frac{1}{2.5 \times 10^{-15} \text{ m}^2} = 0.4 \times 10^{15} \text{ m}^{-2} = 4 \times 10^{14} \text{ m}^{-2}

1λ12=1(1×10−7 m)2=11×10−14 m2=1×1014 m−2\frac{1}{\lambda_1^2} = \frac{1}{(1 \times 10^{-7} \text{ m})^2} = \frac{1}{1 \times 10^{-14} \text{ m}^2} = 1 \times 10^{14} \text{ m}^{-2}

Now, substitute these values back into the ΔKE\Delta KE equation:

ΔKE=(2×10−37 J.m2)×(4×1014 m−2−1×1014 m−2)\Delta KE = (2 \times 10^{-37} \text{ J.m}^2) \times (4 \times 10^{14} \text{ m}^{-2} - 1 \times 10^{14} \text{ m}^{-2})

ΔKE=(2×10−37 J.m2)×(3×1014 m−2)\Delta KE = (2 \times 10^{-37} \text{ J.m}^2) \times (3 \times 10^{14} \text{ m}^{-2})

ΔKE=6×10−23 J\Delta KE = 6 \times 10^{-23} \text{ J}

Finally, convert the energy from Joules to electron volts (eV):

ΔKE(eV)=6×10−23 J1.6×10−19 J/eV\Delta KE (\text{eV}) = \frac{6 \times 10^{-23} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}}

ΔKE(eV)=61.6×10−23−(−19) eV\Delta KE (\text{eV}) = \frac{6}{1.6} \times 10^{-23 - (-19)} \text{ eV}

ΔKE(eV)=3.75×10−4 eV\Delta KE (\text{eV}) = 3.75 \times 10^{-4} \text{ eV}