If the maximum velocity with which an electro | vedclass

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

Question

(B) Given,maximum velocity $v_{\max} = 4 	imes 10^8 \,cm/sec = 4 	imes 10^6 \,m/sec$.Mass of electron $m = 9 	imes 10^{-31} \,kg$.Charge of electro

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) Given,maximum velocity $v_{\max} = 4 	imes 10^8 \,cm/sec = 4 	imes 10^6 \,m/sec$.Mass of electron $m = 9 	imes 10^{-31} \,kg$.Charge of electron $e = 1.6 	imes 10^{-19} \,C$.The maximum kinetic energy is given by $K_{\max} = \frac{1}{2} m v_{\max}^2$.$K_{\max} = \frac{1}{2} 	imes (9 	imes 10^{-31} \,kg) 	imes (4 	imes 10^6 \,m/sec)^2$.$K_{\max} = 0.5 	imes 9 	imes 10^{-31} 	imes 16 	imes 10^{12} = 72 	imes 10^{-19} \,J$.To convert this into electron-volts $(eV)$,divide by $e$:$K_{\max} = \frac{72 	imes 10^{-19} \,J}{1.6 	imes 10^{-19} \,C} = 45 \,eV$.The stopping potential $V_0$ is defined as $K_{\max} = e V_0$,so $V_0 = \frac{K_{\max}}{e} = 45 \,V$.

If the maximum velocity with which an electron can be emitted from a photocell is $4 \times 10^8 \,cm/sec$,the stopping potential is ............ $V$ (mass of electron $= 9 \times 10^{-31} \,kg$).

Similar Questions

If $a_1, a_2, \dots, a_n$ are positive real numbers whose product is a fixed number $c$,then the minimum value of $a_1 + a_2 + \dots + a_{n-1} + 2a_n$ is

In a quadrilateral $ABCD$,the point $P$ divides $DC$ in the ratio $1:2$ and $Q$ is the midpoint of $AC$. If $\overrightarrow{AB}+2\overrightarrow{AD}+\overrightarrow{BC}-2\overrightarrow{DC}=k\overrightarrow{PQ}$,then $k$ is equal to

What are the oxidation states of iron and chromium in chromite ore,respectively?

What is $Y$ in the following reaction?
$C_2H_5I + NaOC_2H_5 \longrightarrow X + NaI$
$X + 2HI \xrightarrow{\Delta} 2Y + H_2O$

Vedclass Test Series

Exam Paper Generator

Online Exam Module