A nucleus with mass number 220 initially at rest emits an $α$ -particle. If the Q value of the reaction is 5.5 MeV. Calculate the kinetic energy of the $α$ -particle

Question

A nucleus with mass number 220 initially rest emits an alpha  particle- If the Q value of the reaction is 5-5 MeV- calculate the kinetic energy of the alpha particle4-4 MeV5-4 MeV5-6 MeV6-5 MeV

Toppr Admin · Accepted Answer

M-x2192-X-x3B1-E-p22mLet Total energy - TT-E-x3B1-EXM-M-x3B1-MXMX-216Conservation of momentump-x3B1-pXm-x3B1-Ealpha-mXEXE-x3B1-mxMTE-x3B1-2162205-5-5-4MeV

A nucleus with mass number $220$ initially at rest emits an $α p a r t i c l e$ . If the $Q$ value of the reaction is $5.5 M e V$ , calculate the kinetic energy of the $α p a r t i c l e$
$4.4 M e V$
$5.4 M e V$
$5.6 M e V$
$6.5 M e V$

$M \to X + α$
$E = \frac{p^{2}}{2 m}$
Let Total energy = $T$
$T = E_{α} + E_{X}$
$M = M_{α} + M_{X}$
$M_{X} = 216$
Conservation of momentum
$p_{α} = p_{X}$
$m_{α} E_{a l p h a} = m_{X} E_{X}$
$E_{α} = \frac{m_{x}}{M} T$
$E_{α} = \frac{216}{220} 5.5 = 5.4 M e V$

A nucleus with mass number 220 initially at rest emits an α particle. If the Q value of the reaction is 5.5MeV, calculate the kinetic energy of the α particle4.4MeV5.4MeV5.6MeV6.5MeV

M→X+αE=p22mLet Total energy = TT=Eα+EXM=Mα+MXMX=216Conservation of momentumpα=pXmαEalpha=mXEXEα=mxMTEα=2162205.5=5.4MeV

A nucleus with mass number $220$ initially at rest emits an $α p a r t i c l e$ . If the $Q$ value of the reaction is $5.5 M e V$ , calculate the kinetic energy of the $α p a r t i c l e$
$4.4 M e V$
$5.4 M e V$
$5.6 M e V$
$6.5 M e V$

$M \to X + α$
$E = \frac{p^{2}}{2 m}$
Let Total energy = $T$
$T = E_{α} + E_{X}$
$M = M_{α} + M_{X}$
$M_{X} = 216$
Conservation of momentum
$p_{α} = p_{X}$
$m_{α} E_{a l p h a} = m_{X} E_{X}$
$E_{α} = \frac{m_{x}}{M} T$
$E_{α} = \frac{216}{220} 5.5 = 5.4 M e V$