An electron is accelerated through a potential difference of 100 V. Calculate: (i) The momentum of the electron. (ii) The de Broglie wavelength associated with the electron. (Given: Mass of electron $m = 9.1 imes 10^{-31} ext{ kg}$, Planck's constant $h = 6.63 imes 10^{-34} ext{ J s}$, charge of electron $e = 1.6 imes 10^{-19} ext{ C}$)

Q: An electron is accelerated through a potential difference of 100 V. Calculate: (i) The momentum of the electron. (ii) The de Broglie wavelength associated with the electron. (Given: Mass of electron $m = 9.1 \times 10^{-31} \text{ kg}$, Planck's constant $h = 6.63 \times 10^{-34} \text{ J s}$, charge of electron $e = 1.6 \times 10^{-19} \text{ C}$)

Given Data: Potential difference ($V$) = $100 \text{ V}$ Mass of electron ($m$) = $9.1 \times 10^{-31} \text{ kg}$ Planck's constant ($h$) = $6.63 \times 10^{-34} \text{ J s}$ Charge of electron ($e$) = $1.6 \times 10^{-19} \text{ C}$ --- (i) Calculation of Momentum ($p$)\nThe kinetic energy acquired by the electron when accelerated through a potential difference $V$ is given by: $$K = eV$$ \nThe relation between momentum ($p$) and kinetic energy ($K$) is: $$p = \sqrt{2mK} = \sqrt{2meV}$| \nSubstituting the given values into the formula: $$p = \sqrt{2 \times (9.1 \times 10^{-31} \text{ kg}) \times (1.6 \times 10^{-19} \text{ C}) \times (100 \text{ V})}$$ $$p = \sqrt{291.2 \times 10^{-50}}$$ $$p = \sqrt{29.12 \times 10^{-49}}$$ $$p = \sqrt{291.2} \times 10^{-25}$| $$p \approx 5.4 \times 10^{-24} \text{ kg m s}^{-1}$| --- (ii) Calculation of de Broglie Wavelength ($\lambda$)\nThe de Broglie wavelength associated with the electron is given by the formula: $$\lambda = \frac{h}{p}$| \nSubstituting the values of $h$ and $p$: $$\lambda = \frac{6.63 \times 10^{-34} \text{ J s}}{5.4 \times 10^{-24} \text{ kg m s}^{-1}}$$ $$\lambda = 1.228 \times 10^{-10} \text{ m}$| $$\lambda \approx 0.123 \text{ nm} \text{ (or } 1.23 \text{ \u00c5)}$$ Final Answer: (i) Momentum of the electron = $5.4 \times 10^{-24} \text{ kg m s}^{-1}$ (ii) de Broglie wavelength = $0.123 \text{ nm}$

Question

An electron is accelerated through a potential difference of 100 V. Calculate: (i) The momentum of the electron. (ii) The de Broglie wavelength associated with the electron. (Given: Mass of electron $m = 9.1 	imes 10^{-31} 	ext{ kg}$, Planck's constant $h = 6.63 	imes 10^{-34} 	ext{ J s}$, charge of electron $e = 1.6 	imes 10^{-19} 	ext{ C}$)

Accepted Answer

Given Data: Potential difference ($V$) = $100 	ext{ V}$ Mass of electron ($m$) = $9.1 	imes 10^{-31} 	ext{ kg}$ Planck's constant ($h$) = $6.63 	imes 10^{-34} 	ext{ J s}$ Charge of electron ($e$) = $1.6 	imes 10^{-19} 	ext{ C}$ --- (i) Calculation of Momentum ($p$)
The kinetic energy acquired by the electron when accelerated through a potential difference $V$ is given by: $$K = eV$$ 
The relation between momentum ($p$) and kinetic energy ($K$) is: $$p = \sqrt{2mK} = \sqrt{2meV}$| 
Substituting the given values into the formula: $$p = \sqrt{2 	imes (9.1 	imes 10^{-31} 	ext{ kg}) 	imes (1.6 	imes 10^{-19} 	ext{ C}) 	imes (100 	ext{ V})}$$ $$p = \sqrt{291.2 	imes 10^{-50}}$$ $$p = \sqrt{29.12 	imes 10^{-49}}$$ $$p = \sqrt{291.2} 	imes 10^{-25}$| $$p \approx 5.4 	imes 10^{-24} 	ext{ kg m s}^{-1}$| --- (ii) Calculation of de Broglie Wavelength ($\lambda$)
The de Broglie wavelength associated with the electron is given by the formula: $$\lambda = \frac{h}{p}$| 
Substituting the values of $h$ and $p$: $$\lambda = \frac{6.63 	imes 10^{-34} 	ext{ J s}}{5.4 	imes 10^{-24} 	ext{ kg m s}^{-1}}$$ $$\lambda = 1.228 	imes 10^{-10} 	ext{ m}$| $$\lambda \approx 0.123 	ext{ nm} 	ext{ (or } 1.23 	ext{ \u00c5)}$$ Final Answer: (i) Momentum of the electron = $5.4 	imes 10^{-24} 	ext{ kg m s}^{-1}$ (ii) de Broglie wavelength = $0.123 	ext{ nm}$