A proton with a kinetic energy of 2.0 , eV mo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The kinetic energy $K = 2.0 \, eV = 2.0 	imes 1.6 	imes 10^{-19} \, J = 3.2 	imes 10^{-19} \, J$.The velocity $v$ is given by $K = \frac{1}{2}m

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) The kinetic energy $K = 2.0 \, eV = 2.0 	imes 1.6 	imes 10^{-19} \, J = 3.2 	imes 10^{-19} \, J$.The velocity $v$ is given by $K = \frac{1}{2}mv^2$,so $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 	imes 3.2 	imes 10^{-19}}{1.6 	imes 10^{-27}}} = \sqrt{4 	imes 10^8} = 2 	imes 10^4 \, m/s$.The pitch $p$ of the helical path is given by $p = (v \cos 	heta) 	imes T$,where $T = \frac{2\pi m}{qB}$ is the time period.Substituting the values: $p = v \cos 60^{\circ} 	imes \frac{2\pi m}{qB}$.$p = (2 	imes 10^4) 	imes \frac{1}{2} 	imes \frac{2\pi 	imes 1.6 	imes 10^{-27}}{1.6 	imes 10^{-19} 	imes (\frac{\pi}{2} 	imes 10^{-3})}$.$p = 10^4 	imes \frac{2 	imes 10^{-27}}{10^{-19} 	imes 0.5 	imes 10^{-3}} = 10^4 	imes 4 	imes 10^{-5} = 0.4 \, m$.Converting to centimeters: $p = 0.4 	imes 100 = 40 \, cm$.

Similar Questions

$A$ particle is projected with a velocity of $10 \ m/s$ along the $y-$axis from the point $(2, 3)$. $A$ uniform magnetic field of $(3\hat{i} + 4\hat{j}) \ T$ exists in the space. What is its speed when the particle passes through the $y-$axis for the third time? (Neglect gravity)

$A$ uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5 \ Wb/m^2$. If a proton having mass $m = 1.7 \times 10^{-27} \ kg$ and charge $q = 1.6 \times 10^{-19} \ C$ moves in this field vertically downwards with energy $5 \ MeV$,then the force acting on it will be

$A$ deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let $r_{d}$ and $r_{\alpha}$ be their respective radii of circular path. The value of $\frac{r_{d}}{r_{\alpha}}$ is equal to

$A$ uniform magnetic field acts at right angles to the direction of motion of electrons. As a result,the electron moves in a circular path of radius $2\, cm$. If the speed of the electrons is doubled,then the radius of the circular path will be.....$cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module