Initially the nucleus of radium – 226 is at rest. It decays due to which and a particle and the nucleus of radon are created. The released energy during the decay is 4.87 Mev, which appears as the kinetic energy of the two resulted particles. $[m_{α} = 4.002 a m u, m_{R n} = 222.017 a m u]$

Kinetic energies of particle & radon nucleus are respectively

0.09 Mev, 4.08 Mev

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4.78 Mev, 0.09Mev

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4.08 Mev, 0.09 Mev

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3.68 Mev, 0.08 Mev

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Solution

The correct options are
C
4.08 Mev, 0.09 Mev

D
3.68 Mev, 0.08 Mev

$P = m v$ and
$\frac{P^{2}}{2 m} = \frac{1}{2} m V^{2} E_{a} = \frac{P^{2}}{2 m} E_{r a d o n} = \frac{P^{2}}{2 M}$
Also
$\frac{P^{2}}{2 m} + \frac{P^{2}}{2 M} = Δ E \Rightarrow P = \sqrt{2 Δ E \frac{m M}{m + M}} m = 4.002 m u, M = 222.017 m u m = 6.64 \times 10^{- 27} k g$

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Similar questions

Q. Initially, the nucleus of

Radium - 226

is at rest. It decays by emitting an

α -

particle, and thus the nucleus of

Radon

is created. The released energy during the decay is

4.87 MeV

, which appears as the kinetic energy of the two resulted particles. The kinetic energy of

α -

particle &

Radon

nucleus are respectively.

[m_{α} = 4.002 u, m_{Rn} = 222.017 u]

Initially the nucleus of radium – 226 is at rest. It decays due to which and a particle and the nucleus of radon are created. The released energy during the decay is 4.87 Mev, which appears as the kinetic energy of the two resulted particles. $[m_{α} = 4.002 a m u, m_{R n} = 222.017 a m u]$