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  • The velocity of an electron is increased by passing it through a potential difference of 100 V. The de-Broglie wavelength associated with it is found to be 0.0123 nm. What would be the de-Broglie w

The velocity of an electron is increased by passing it through a potential difference of 100 V. The de-Broglie wavelength associated with it is found to be 0.0123 nm. What would be the de-Broglie wavelength of an alpha particle accelerated through the same potential difference? (Take the mass of alpha particle to be 6.64\times10^{-27}kg and its charge to be 2e)

Option: 1

0.025 nm


Option: 2

 0.0316 nm


Option: 3

0.1 nm


Option: 4

0.2 nm


Answers (1)

best_answer

We can use the de Broglie wavelength formula to solve this problem:

\lambda= \frac{h}{p}

 

where \lambda is the de Broglie wavelength, h is Planck's constant, and p is the momentum of the particle.
We know that the electron has been accelerated through a potential difference of 100 V. The energy gained by the electron is given by:

 

E= qV

where E is the energy gained, q is the charge of the electron, and V is the potential difference.
Substituting the values, we get:

E= \left ( 1.6\times 10^{-19}C \right )\times\left ( 100 V \right )= 1.6\times 10^{-17}J
The momentum of the electron can be calculated using:

E=\frac{p^{2}}{2m}
where m is the mass of the electron. Substituting the values, we get:

p=\sqrt{2 m E}=5.25 \times 10^{-24} \mathrm{~kg} \mathrm{~m} / \mathrm{s}
Now, we can calculate the de Broglie wavelength of the electron using:

\lambda=\frac{h}{p}=\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{5.25 \times 10^{-24} \mathrm{~kg} \mathrm{~m} / \mathrm{s}}=0.0123 \mathrm{~nm}
For the alpha particle, we can use the same approach. The energy gained by the alpha particle is:

E=q V=\left(2 \times 1.6 \times 10^{-19} \mathrm{C}\right) \times(100 \mathrm{~V})=3.2 \times 10^{-17} \mathrm{~J}

The momentum of the alpha particle can be calculated using:

p=\sqrt{2 m E}=\sqrt{2 \times 6.64 \times 10^{-27} \mathrm{~kg} \times 3.2 \times 10^{-17} \mathrm{~J}}=2.09 \times10^{-21}kgm/s

Now, we can calculate the de Broglie wavelength of the alpha particle using:

\lambda=\frac{h}{p}=\frac{6.626 \times 10^{-34} \mathrm{Js}}{2.09 \times 10^{-21} \mathrm{kgm} / \mathrm{s}} \approx 0.0316 \mathrm{~nm}

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Gaurav

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