1
JEE Main 2023 (Online) 31st January Evening Shift
+4
-1
The radius of electron's second stationary orbit in Bohr's atom is R. The radius of 3rd orbit will be
A
2.25R
B
$3 \mathrm{R}$
C
$\frac{\mathrm{R}}{3}$
D
$9 \mathrm{R}$
2
JEE Main 2023 (Online) 31st January Morning Shift
+4
-1
Out of Syllabus

A free neutron decays into a proton but a free proton does not decay into neutron. This is because

A
neutron is an uncharged particle
B
neutron has larger rest mass than proton
C
neutron is a composite particle made of a proton and an electron
D
proton is a charged particle
3
JEE Main 2023 (Online) 30th January Evening Shift
+4
-1
Out of Syllabus
Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$

Assertion A: The nuclear density of nuclides ${ }_{5}^{10} \mathrm{~B},{ }_{3}^{6} \mathrm{Li},{ }_{26}^{56} \mathrm{Fe},{ }_{10}^{20} \mathrm{Ne}$ and ${ }_{83}^{209} \mathrm{Bi}$ can be arranged as $\rho_{\mathrm{Bi}}^{\mathrm{N}}>\rho_{\mathrm{Fe}}^{\mathrm{N}}>\rho_{\mathrm{Ne}}^{\mathrm{N}}>\rho_{\mathrm{B}}^{\mathrm{N}}>\rho_{\mathrm{Li}}^{\mathrm{N}}$

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 statements, choose the correct answer from the options given below
A
${Both ~\mathbf{A}}$ and $\mathbf{R}$ are true and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$
B
Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is NOT the correct explanation of $\mathbf{A}$
C
$\mathbf{A}$ is false but $\mathbf{R}$ is true
D
$\mathbf{A}$ is true but $\mathbf{R}$ is false
4
JEE Main 2023 (Online) 30th January Morning Shift
+4
-1

Speed of an electron in Bohr's $$7^{\text {th }}$$ orbit for Hydrogen atom is $$3.6 \times 10^{6} \mathrm{~m} / \mathrm{s}$$. The corresponding speed of the electron in $$3^{\text {rd }}$$ orbit, in $$\mathrm{m} / \mathrm{s}$$ is :

A
$$\left(1.8 \times 10^{6}\right)$$
B
$$\left(7.5 \times 10^{6}\right)$$
C
$$\left(8.4 \times 10^{6}\right)$$
D
$$\left(3.6 \times 10^{6}\right)$$
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