The mass of an $\alpha$-particle is
The mass of an $\alpha$-particle is
- [AIPMT 1992]
- A
Less than the sum of masses of two protons and two neutrons
- B
Equal to mass of four protons
- C
Equal to mass of four neutrons .
- D
Equal to sum of masses of two protons and two neutrons
Similar Questions
An alpha nucleus of energy $\frac{1}{2}m{v^2}$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to
An alpha nucleus of energy $\frac{1}{2}m{v^2}$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason $R$. Assertion A : The nuclear density of nuclides ${ }_5^{10} B ,{ }_3^6 Li ,{ }_{26}^{56} Fe ,{ }_{10}^{20} Ne$ and ${ }_{83}^{209} Bi$ can be arranged as $\rho_{ Bi }^{ N }>\rho_{ Fe }^{ N }>\rho_{ Ne }^{ N }>\rho_{ B }^{ N }>\rho_{ Li }^{ N }$.
Reason $R$ : The radius $R$ of nucleus is related to its mass number $A$ as $R=R_0 A^{1 / 3}$, where $R_0$ is a constant.
In the light of the above statement, choose the correct answer from the options given below :
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason $R$. Assertion A : The nuclear density of nuclides ${ }_5^{10} B ,{ }_3^6 Li ,{ }_{26}^{56} Fe ,{ }_{10}^{20} Ne$ and ${ }_{83}^{209} Bi$ can be arranged as $\rho_{ Bi }^{ N }>\rho_{ Fe }^{ N }>\rho_{ Ne }^{ N }>\rho_{ B }^{ N }>\rho_{ Li }^{ N }$.
Reason $R$ : The radius $R$ of nucleus is related to its mass number $A$ as $R=R_0 A^{1 / 3}$, where $R_0$ is a constant.
In the light of the above statement, choose the correct answer from the options given below :
- [JEE MAIN 2023]
The charge density in a nucleus varies with distance from the centre of the nucleus according to the curve in Fig.
The charge density in a nucleus varies with distance from the centre of the nucleus according to the curve in Fig.
Assertion $(A):$ Forces acting between proton-protn $\left(f_{p p}\right)$, proton-neutron $\left(f_{p p}\right)$ and neutron-neutron $\left(f_{n n}\right)$ are such that $f_{p p} < f_{p n}=f_{n n}$
Reason $(R):$ Electrostatic force of repulsion between two protons reduces net nuclear forces between them.
Assertion $(A):$ Forces acting between proton-protn $\left(f_{p p}\right)$, proton-neutron $\left(f_{p p}\right)$ and neutron-neutron $\left(f_{n n}\right)$ are such that $f_{p p} < f_{p n}=f_{n n}$
Reason $(R):$ Electrostatic force of repulsion between two protons reduces net nuclear forces between them.
- [AIIMS 2015]
Which of the following isotopes is used for the treatment of cancer
Which of the following isotopes is used for the treatment of cancer