[Take the mass of electron to be $9 \times 10^{-31}\text{ kg}$, charge of electron $= 1.6 \times 10^{-19}\text{ C}$ and $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$]
Physics Problem: Electron Orbit Radius Calculation
This problem involves calculating the radius of an electron's orbit in a Hydrogen atom, given its velocity and other fundamental constants.
Physics Principle Applied
For a stable circular orbit, the electrostatic force of attraction between the nucleus (proton) and the electron provides the necessary centripetal force for the electron's motion.
Formula Derivation
The electrostatic force ($F_e$) is given by Coulomb's law, and the centripetal force ($F_c$) is given by:
- $F_e = k \frac{e^2}{r^2}$
- $F_c = \frac{m_e v^2}{r}$
Equating these forces ($F_e = F_c$):
$ k \frac{e^2}{r^2} = \frac{m_e v^2}{r} $
Solving for the radius ($r$):
$ r = \frac{k e^2}{m_e v^2} $
Calculation Steps
Given values:
- Velocity, $v = \sqrt{25.6} \times 10^5 \text{ ms}^{-1}$
- Mass of electron, $m_e = 9 \times 10^{-31} \text{ kg}$
- Charge of electron, $e = 1.6 \times 10^{-19} \text{ C}$
- Electrostatic constant, $k = 9 \times 10^9 \text{ N m}^2\text{ C}^{-2}$
First, calculate $v^2$:
$ v^2 = (\sqrt{25.6} \times 10^5)^2 = 25.6 \times (10^5)^2 = 25.6 \times 10^{10} \text{ m}^2\text{s}^{-2} $
Next, calculate $e^2$:
$ e^2 = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \text{ C}^2 $
Now, substitute these values into the formula for $r$:
$ r = \frac{(9 \times 10^9 \text{ N m}^2\text{ C}^{-2}) \times (2.56 \times 10^{-38} \text{ C}^2)}{(9 \times 10^{-31} \text{ kg}) \times (25.6 \times 10^{10} \text{ m}^2\text{s}^{-2})} $
Simplify the expression:
$ r = \frac{9 \times 2.56 \times 10^{(9 - 38)}}{9 \times 25.6 \times 10^{(-31 + 10)}} $
$ r = \frac{2.56 \times 10^{-29}}{25.6 \times 10^{-21}} $
$ r = \frac{2.56}{25.6} \times 10^{-29 - (-21)} $
$ r = 0.1 \times 10^{-8} \text{ m} $
$ r = 1 \times 10^{-9} \text{ m} $
Result Analysis
The calculated radius is $r = 1 \times 10^{-9} \text{ m}$.
The problem states the radius is $x \times 10^{-9} \text{ m}$. By comparing the calculated value with the given format, we find:
$ x = 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 :- An unknown nucleus has a nuclear density of $2.29 \times 10^{17} \text{ kg/m}^3$ and mass of $19.926 \times 10^{-27} \text{ kg}$. Its mass number A is approximately :
(Take $R_0 = 1.2 \times 10^{-15} \text{ m}$, $4\pi = 12.56$) - 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 $= 9 \times 10^{-31}\text{ kg}$) - 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 $6.6 \times 10^{-34} \text{ J s}$)
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 ______ $\Omega$.
- 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 $= 6.6 \times 10^{-34}\text{ J.s}$)
- The average energy released per fission for the nucleus of ${}^{235}_{92}\text{U}$ is $190 \text{ MeV}$. When all the atoms of $47 \text{ g}$ pure ${}^{235}_{92}\text{U}$ undergo fission process, the energy released is $\alpha \times 10^{23} \text{ MeV}$. The value of $\alpha$ is ________.
(Avogadro Number = $6 \times 10^{23} \text{ per mole}$) - The energy released when $\frac{7}{17.13} \text{ kg}$ of ${}^7_3\text{Li}$ is converted into ${}^4_2\text{He}$ by proton bombardment is $\alpha \times 10^{32} \text{ eV}$. The value of $\alpha$ is _________. (Nearest integer)
(Mass of ${}^7_3\text{Li} = 7.0183 \text{ u}$, mass of ${}^4_2\text{He} = 4.004 \text{ u}$, mass of proton = $1.008 \text{ u}$ and $1 \text{ u} = 931 \text{ MeV/}c^2$ and Avogadro number $= 6.0 \times 10^{23}$) - A nuclear physicist performed an experiment on the scattering of alpha particles by thin metal foil of gold. The alpha particles were accelerated up to kinetic energy $3.2$ Mev. The estimation of the radius of the nucleus of gold element $(Z = 79)$ by him is $\times 10^{-16}$ m
(Take $\frac{1}{4\pi \epsilon_0} = 9\times10^9 \frac{Nm^2}{C^2}$)
