A scientist claims to have a perfect technique in which he can spontaneously convert an electron com

Gujarati

Hindi

13.Nuclei

medium

A scientist claims to have a perfect technique in which he can spontaneously convert an electron completely into energy in the laboratory without any other material required. What is the conclusion about this claim from our current understanding of physics?

This is possible because Einstein's equation says that mass and energy are equivalent... it is just very difficult to achieve with electrons

This is possible and it is done all the time in the high-energy physics labs.

The scientist is almost correct... except that in converting the electron to energy, an electron's anti-particle is produced in the process as well

This is not possible because charge conservation would be violated.

Solution

Charge is conserved. In order to fully convert an electron into energy, a positron (the electron's antiparticle must be involved). That is, electron + positron -> energy, $NOT$ electron -> positron + energy.]

Standard 12

Physics

When the atomic number $A$ of the nucleus increases

medium

The mass of a neutron is the same as that of

easy

A nucleus disintegrates into two nuclear parts, in such a way that ratio of their nuclear sizes is $1: 2^{1 / 3}$.Their respective speed have a ratio of $n: 1$. The value of $n$ is $…..$.

medium

(JEE MAIN-2023)

Size of nucleus is of the order of

medium

The radius of $R$ of a nucleus of mass number $A$ can be estimated by the formula $R =\left(1.3 \times 10^{-15}\right) A ^{1 / 3}\, m .$ It follows that the mass density of a nucleus is of the order of

$\left( M _{\text {prot. }} \cong M _{\text {neut. }}=1.67 \times 10^{-27} kg \right)$

medium

(JEE MAIN-2020)

A scientist claims to have a perfect technique in which he can spontaneously convert an electron completely into energy in the laboratory without any other material required. What is the conclusion about this claim from our current understanding of physics?

Solution

Similar Questions

When the atomic number $A$ of the nucleus increases

The mass of a neutron is the same as that of

A nucleus disintegrates into two nuclear parts, in such a way that ratio of their nuclear sizes is $1: 2^{1 / 3}$.Their respective speed have a ratio of $n: 1$. The value of $n$ is $…..$.

Size of nucleus is of the order of

The radius of $R$ of a nucleus of mass number $A$ can be estimated by the formula $R =\left(1.3 \times 10^{-15}\right) A ^{1 / 3}\, m .$ It follows that the mass density of a nucleus is of the order of $\left( M _{\text {prot. }} \cong M _{\text {neut. }}=1.67 \times 10^{-27} kg \right)$

Start a Free Trial Now

The radius of $R$ of a nucleus of mass number $A$ can be estimated by the formula $R =\left(1.3 \times 10^{-15}\right) A ^{1 / 3}\, m .$ It follows that the mass density of a nucleus is of the order of

$\left( M _{\text {prot. }} \cong M _{\text {neut. }}=1.67 \times 10^{-27} kg \right)$