(Take $R_0 = 1.2 \times 10^{-15} \text{ m}$, $4\pi = 12.56$)
Nuclear Density Calculation Steps
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 Formula
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 ($1.2 \times 10^{-15} \text{ m}$).
Combining Formulas
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} $Calculating Mass Number A
Rearrange the formula to solve for the mass number ($A$):
$ A = \frac{m}{\rho \frac{4}{3}\pi R_0^3} $Given values:
- Mass ($m$) = $19.926 \times 10^{-27} \text{ kg}$
- Nuclear density ($\rho$) = $2.29 \times 10^{17} \text{ kg/m}^3$
- $R_0 = 1.2 \times 10^{-15} \text{ m}$
- $4\pi \approx 12.56$
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 $Conclusion
The calculated mass number $A$ is approximately 12.026. Rounding to the nearest integer gives $A = 12$. This corresponds to Option 1.
Similar Questions
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 :- For a metal of work function $6.6 \text{ eV}$, which of the following wavelengths of incident radiation does not give rise to the photoelectric effect?
(Take Planck's constant as $6.6 \times 10^{-34} \text{ J s}$) 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 Geiger-Marsden experiment, the number of scattered $\alpha$-particles $N(\theta)$ is plotted as a function of scattering angle $\theta$. Which of the following options represents the correct plot ?
- A beam of light falls on a metal surface such that photo-electrons are generated. If power of the light source starts to decrease linearly with time $t$, then variation of the photocurrent $I$ and magnitude of the stopping potential $|V|$ with time is best represented by :
- A ray of light with wavelength $\lambda$ is incident on three different photo-electric cells namely 1, 2 and 3. The threshold wavelength of these photo-electric cells are $\lambda_1$, $\lambda_2$, and $\lambda_3$, respectively and the magnitude of stopping potentials of these cells are $V_1$, $V_2$ and $V_3$, respectively. The relation between $\lambda$ and threshold wavelengths are $\lambda_1 < \lambda$, $\lambda_2 > \lambda$ and $\lambda_3 \gg \lambda$. The correct option is :
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 :
- Consider the following nuclear reaction :
$$^{238}\text{U} \rightarrow ^{234}\text{Th} + ^{4}\text{He}$$
Take masses of $^{238}\text{U}$, $^{234}\text{Th}$ and $^{4}\text{He}$ as $238.050\text{ u}$, $234.043\text{ u}$ and $4.003\text{ u}$, respectively. The Q value for the reaction, in keV, is :
[Given : $1\text{ u} = 931.5\text{ MeV c}^{-2}$] - A photon and an electron, each of $20\text{ eV}$ energy, move in free space. The ratio of linear momentum of electron $p_e$ to that of photon $p_{ph}$, $\frac{p_e}{p_{ph}}$ is :
(Take speed of light $= 3 \times 10^8\text{ ms}^{-1}$, charge of electron $= -1.6 \times 10^{-19}\text{ C}$ and mass of electron $= 9 \times 10^{-31}\text{ kg}$)
Important Questions from Modern Physics
- A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W, is operated at 15 V. The approximate value of protective resistance in this circuit is ______ $\Omega$.
The binding energy for the following nuclear reactions are expressed in MeV.
${}_2\text{He}^3 + {}_0\text{n}^1 \rightarrow {}_2\text{He}^4 + 20 \text{ MeV}$
${}_2\text{He}^4 + {}_0\text{n}^1 \rightarrow {}_2\text{He}^5 - 0.9 \text{ MeV}$
If $\text{X}_3, \text{X}_4, \text{X}_5$ denote the stability of ${}_2\text{He}^3, {}_2\text{He}^4$ and ${}_2\text{He}^5$, respectively, then the correct order is :Identify the correct truth table of the given logic circuit.
- When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is $3.2 \text{ V}$. If a second light having wavelength twice of first light is used, the stopping potential drops to $0.7 \text{ V}$. The wavelength of first light is _______ m.
($h = 6.63 \times 10^{-34} \text{ J.s}, e = 1.6 \times 10^{-19} \text{ C}, c = 3 \times 10^8 \text{ m/s}$) - A particle having electric charge $3 \times 10^{-19}\text{ C}$ and mass $6 \times 10^{-27}\text{ kg}$ is accelerated by applying an electric potential of $1.21\text{ V}$. Wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12}\text{ m}$. The value of $\alpha$ is __________. (Take Planck's constant $= 6.6 \times 10^{-34}\text{ J.s}$)
