(Take $R_0 = 1.2 \times 10^{-15} \text{ m}$, $4\pi = 12.56$)
To find the approximate mass number (A) of the nucleus, we use the relationship between nuclear density ($\rho$), the mass of the nucleus ($m$), and the nuclear radius ($R$).
Nuclear density is defined as mass per unit volume:
$ \rho = \frac{m}{V} $The volume ($V$) of a nucleus is approximated as a sphere with radius $R$:
$ V = \frac{4}{3}\pi R^3 $The nuclear radius ($R$) is related to the mass number ($A$) by the empirical formula:
$ R = R_0 A^{1/3} $Where $R_0$ is a constant ($1.2 \times 10^{-15} \text{ m}$).
Substituting the expression for $R$ into the volume formula:
$ V = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A $Now, substitute this volume into the density formula:
$ \rho = \frac{m}{\frac{4}{3}\pi R_0^3 A} $Rearrange the formula to solve for the mass number ($A$):
$ A = \frac{m}{\rho \frac{4}{3}\pi R_0^3} $Given values:
Calculate the term $\frac{4}{3}\pi R_0^3$:
$ \frac{4}{3}\pi R_0^3 \approx \frac{1}{3} \times (12.56) \times (1.2 \times 10^{-15} \text{ m})^3 $ $ \frac{4}{3}\pi R_0^3 \approx \frac{12.56}{3} \times (1.728 \times 10^{-45} \text{ m}^3) $ $ \frac{4}{3}\pi R_0^3 \approx 4.1867 \times 1.728 \times 10^{-45} \text{ m}^3 \approx 7.235 \times 10^{-45} \text{ m}^3 $Now substitute all values into the formula for $A$:
$ A = \frac{19.926 \times 10^{-27} \text{ kg}}{(2.29 \times 10^{17} \text{ kg/m}^3) \times (7.235 \times 10^{-45} \text{ m}^3)} $ $ A = \frac{19.926 \times 10^{-27}}{2.29 \times 7.235 \times 10^{17 - 45}} $ $ A = \frac{19.926 \times 10^{-27}}{16.568 \times 10^{-28}} $ $ A = \frac{19.926}{1.6568} \approx 12.026 $The calculated mass number $A$ is approximately 12.026. Rounding to the nearest integer gives $A = 12$. This corresponds to Option 1.
Match List I with List II :
| List I | List II |
| A. $E = h\nu$ | I. de Broglie wavelength |
| B. Interference | II. Particle nature of light |
| C. $\lambda = h/p$ | III. Wave nature of light |
| D. Compton effect | IV. Energy of photon |
Choose the correct answer from the options given below :
In the first excited state of hydrogen atom, the energy of its electron is $-3.4 \text{ eV}$. The radial distance of the electron from the hydrogen nucleus in this case is approximately :
(Take $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}, \text{ e} = 1.6 \times 10^{-19} \text{ C}$ and $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$)
Four statements are given (A is mass number) :
A. The volume of a nucleus is proportional to $A^{1/3}$.
B. The volume of a nucleus is proportional to A.
C. The difference in mass of an atom and its nucleus is called the mass defect.
D. The difference in mass of a nucleus and its constituent nucleons is called the mass defect.
Choose the correct answer from the options given below :
An ideal Zener diode with breakdown voltage of $-3\text{ V}$ is reverse biased with a negative input voltage $V_i = -5\text{ V}$. The magnitude of voltage difference between points B and A is :
Assuming in forward bias condition there is a voltage drop of $0.7 \text{ V}$ across a silicon diode, the current through diode $D_1$ in the circuit is ________$\text{mA}$.
(Assume all diodes in the given circuit are identical)

Find the correct combination of A, B, C and D inputs which can cause the LED to glow.
