Q.48. A particle having electric charge C and mass kg is accelerated by applying an electric poten

Q: What is the correct answer for JEE Main Physics Q48?

The correct answer is 10. Charge of particle $q = 3 \times 10^{-19}$ C.Mass of particle $m = 6 \times 10^{-27}$ kg.Potential difference $V = 1.21$ V.Planck's constant $h = 6.6 \times 10^{-34}$ J.s.The kinetic energy gained by the particle when accelerated through a potential difference V is $K = qV$.The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$, where $p$ is the momentum.We know $K = \frac{p^2}{2m

Q: Q.48. A particle having electric charge C and mass kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is m. The value of is…

The correct answer is 10. Charge of particle $q = 3 \times 10^{-19}$ C.Mass of particle $m = 6 \times 10^{-27}$ kg.Potential difference $V = 1.21$ V.Planck's constant $h = 6.6 \times 10^{-34}$ J.s.The kinetic energy gained by the particle when accelerated through a potential difference V is $K = qV$.The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$, where $p$ is the momentum.We know $K = \frac{p^2}{2m

Question

Q.48. A particle having electric charge  C and mass  kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is  m. The value of  is .......... (Take Planck's constant =  J.s)

FirstInTest · Accepted Answer

Charge of particle $q = 3 	imes 10^{-19}$ C.Mass of particle $m = 6 	imes 10^{-27}$ kg.Potential difference $V = 1.21$ V.Planck's constant $h = 6.6 	imes 10^{-34}$ J.s.The kinetic energy gained by the particle when accelerated through a potential difference V is $K = qV$.The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$, where $p$ is the momentum.We know $K = \frac{p^2}{2m}$, so $p = \sqrt{2mK}$.Therefore, $\lambda = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$.Substitute the given values:$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{2 	imes (6 	imes 10^{-27}) 	imes (3 	imes 10^{-19}) 	imes 1.21}}$$$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{36 	imes 10^{-46} 	imes 1.21}}$$$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{36 	imes 1.21 	imes 10^{-46}}}$$$$\lambda = \frac{6.6 	imes 10^{-34}}{6 	imes 1.1 	imes 10^{-23}}$$$$\lambda = \frac{6.6 	imes 10^{-34}}{6.6 	imes 10^{-23}}$$$$\lambda = 1 	imes 10^{-11} 	ext{ m}$$We are given that $\lambda = \alpha 	imes 10^{-12}$ m.So, $1 	imes 10^{-11} = \alpha 	imes 10^{-12}$.$10 	imes 10^{-12} = \alpha 	imes 10^{-12}$.Therefore, $\alpha = 10$.

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