The Correct Answer is Option (A)

The energy levels of a one-electron ion can be described by the formula:

$E_n=-\frac{Z^2}{n^2}\times E_0$

where $E_n$ is the energy of the nth level, Z is the atomic number (number of protons), n is the principal quantum number, and $E_0$is the ground state energy of the hydrogen atom (-13.6 eV).

For the ${\mathrm{He}}^{+}$ ion, the atomic number Z is 2 (since helium has 2 protons). We are looking for the energy of the first excited state, which corresponds to n = 2. Plugging these values into the formula, we get:

$E_2=-\frac{2^2}{2^2}\times (-13.6\mathrm{eV})=-13.6\mathrm{eV}$

So, the energy of the ${\mathrm{He}}^{+}$ion in its first excited state is$-13.6\mathrm{eV}$.

The Correct Answer is Option (A)

According to Bohr's model of the hydrogen atom, the angular momentum of an electron in an orbit is an integral multiple of Planck's constant divided by $2\pi$ (or $h/2\pi$, where $h$ is the Planck's constant). This can be expressed as:

$L=n\frac{h}{2\pi }$

where $n$ is the principal quantum number or the orbit number.

So, for the first orbit ($n=1$), the angular momentum $L_1$ is:

$L_1=1\times \frac{h}{2\pi }=\frac{h}{2\pi }$

And for the second orbit ($n=2$), the angular momentum $L_2$is:

$L_2=2\times \frac{h}{2\pi }=\frac{h}{\pi }$

The change in angular momentum when moving from the first to the second orbit is the difference between $L_2$ and $L_1$:

$\mathrm{\Delta }L=L_2-L_1=\frac{h}{\pi }-\frac{h}{2\pi }=\frac{h}{2\pi }$

Since $\frac{h}{2\pi }$ is equal to the initial angular momentum $L_1$, the change in angular momentum when the electron moves to the second orbit is $L$.

Therefore, the correct answer is $L$.

The Correct Answer is Option (A)

According to Bohr's quantization of angular momentum, the angular momentum $L$ of a particle in a circular orbit is given by:

$L=n\hslash$

Where $n$ is an integer and $\hslash$ is the reduced Planck's constant. The angular momentum $L$ can also be expressed as:

$L=mvr$

Where $m$ is the mass of the particle,$v$is its linear velocity, and $r$is the radius of the orbit.

Now, we are given the potential energy $U=\frac{1}{2}m{\omega }^2{r}^2$. Since the particle is in a circular orbit, its centripetal force is provided by the gradient of the potential energy:

$m\frac{v^2}{r}=-\frac{\mathrm{d}U}{\mathrm{d}r}=-m{\omega }^{2}r$

We can simplify this equation to get the relation between $v$ and $r$:

$v^2={\omega }^2{r}^2$

Now, let's combine the equations for angular momentum and the relation between $v$ and $r$:

$n\hslash =mvr=m\sqrt{{\omega }^2{r}^2}r=m\omega r^2$

We can now solve for the radius $r$in terms of$n$:

$r^2=\frac{n\hslash }{m\omega }$

Taking the square root of both sides, we get:

$r\propto \sqrt{n}$

So, the radius of the $n^{\text{th}}$ orbit is proportional to $\sqrt{n}$.

The Correct Answer is Option (A)

$r\propto \frac{n^2}{Z}$

$\begin{array}{rl}& \frac{r_{2nd}}{r_{3\mathrm{rd}}}={\left(\frac{n_2}{n_3}\right)}^2\\ \\ & \Rightarrow \frac{R}{r_{3rd}}={\left(\frac{2}{3}\right)}^2\\ \\ & \Rightarrow r_{3\mathrm{rd}}=\frac{9}{4}R\\ \\ & =2.25R\end{array}$

The Correct Answer is Option (C)

$v\alpha \frac{z}{n}$

$\frac{v_1}{v_2}=\left(\frac{n_2}{n_1}\right)$

$\Rightarrow \frac{3.6\times {10}^6}{v_2}=\frac{3}{7}$

$\Rightarrow v_2=\frac{7}{3}\times 3.6\times {10}^6$ m/s

$=8.4\times {10}^6$ m/s

The Crrect Answer is

The formula for the magnetic field due to a moving charge is given by:

$\mathbf{B}=\frac{{\mu }_0}{4\pi }\frac{qv\sin \theta }{r^2}$

where ${\mu }_0$ is the permeability of free space, $q$ is the charge of the moving particle, $v$ is the speed of the particle, $\theta$ is the angle between the velocity vector and the position vector from the particle to the point where we want to calculate the magnetic field, and $r$ is the distance between the particle and the point where we want to calculate the magnetic field.

In this case, we're interested in the magnetic field produced by the electron moving in a circular orbit around the nucleus of a hydrogen atom. Since the orbit is circular, the angle between the velocity vector and the position vector is 90 degrees, so $\sin \theta =1$. We can substitute the known values into the formula to find the magnetic field:

$\mathbf{B}=\frac{{\mu }_0}{4\pi }\frac{qv\sin \theta }{r^2}=\frac{{\mu }_0}{4\pi }\frac{ev}{r^2}$

where $e$ is the charge of an electron. We know that the radius of the orbit is $0.52{\mathrm{A}}^{\circ }$, which is equivalent to $0.52\times {10}^{-10}\mathrm{m}$.

Substituting the values, we get:

$\mathbf{B}=\frac{{\mu }_0}{4\pi }\frac{ev}{r^2}=\frac{{10}^{-7}\times 1.6\times {10}^{-19}\times 6.76\times {10}^6}{0.52\times 0.52\times {10}^{-20}}=40\mathrm{T}$

This means that the magnetic field produced by the electron moving in a circular orbit around the nucleus of a hydrogen atom is 40 tesla, which is an incredibly strong magnetic field.

The Crrect Answer is

To find the value of $x$, we need to first determine the expressions for the radii of the specified orbits for ${\mathrm{He}}^{+}$ and ${\mathrm{Be}}^{3+}$ according to Bohr's model. The radius of an orbit in a hydrogen-like atom (an atom with only one electron) is given by:

$r_n=\frac{n^2\cdot h^2\cdot {ϵ}_0}{\pi \cdot Z\cdot e^2\cdot m_e}$

Where:

• $r_n$ is the radius of the nth orbit
• $n$ is the principal quantum number (orbit number)
• $h$ is the Planck's constant
• ${ϵ}_0$ is the vacuum permittivity
• $Z$ is the atomic number (number of protons in the nucleus)
• $e$ is the elementary charge
• $m_e$ is the mass of the electron
• $\pi$ is the mathematical constant pi

In this problem, we are looking at the 2nd orbit of ${\mathrm{He}}^{+}$ (which has an atomic number $Z=2$) and the 4th orbit of ${\mathrm{Be}}^{3+}$ (which has an atomic number $Z=4$). Let's calculate the radii for these orbits:

For the 2nd orbit of ${\mathrm{He}}^{+}$ ($n_1=2$ and $Z_1=2$):

$r_1=\frac{n_1^2\cdot h^2\cdot {ϵ}_0}{\pi \cdot Z_1\cdot e^2\cdot m_e}$

For the 4th orbit of ${\mathrm{Be}}^{3+}$ ($n_2=4$ and $Z_2=4$):

$r_2=\frac{n_2^2\cdot h^2\cdot {ϵ}_0}{\pi \cdot Z_2\cdot e^2\cdot m_e}$

We are asked to find the ratio $\frac{r_2}{r_1}$, which is equal to $x:1$:

$\frac{r_2}{r_1}=\frac{\frac{n_2^2\cdot h^2\cdot {ϵ}_0}{\pi \cdot Z_2\cdot e^2\cdot m_e}}{\frac{n_1^2\cdot h^2\cdot {ϵ}_0}{\pi \cdot Z_1\cdot e^2\cdot m_e}}$

By simplifying the expression, we get:

$\frac{r_2}{r_1}=\frac{n_2^2\cdot Z_1}{n_1^2\cdot Z_2}=\frac{4^2\cdot 2}{2^2\cdot 4}$

Now we can calculate the value of $x$:

$x=\frac{r_2}{r_1}=\frac{16\cdot 2}{4\cdot 4}=\frac{32}{16}=2$

Therefore, the value of $x$ in the ratio $\frac{r_2}{r_1}=x:1$ is $\boxed{{\displaystyle 2}}$.

The Crrect Answer is

The binding energy of an electron in a hydrogen atom is given by the formula:

$E=\frac{k{e}^2}{2r}$

where:

• $E$ is the energy of the electron,
• $k$ is Coulomb's constant (Nm$9\times {10}^9{\text{Nm}}^2/{\text{C}}^2$),
• $e$ is the charge of the electron ($1.6\times {10}^{-19}\text{C}$), and
• math xmlns="http://www.w3.org/1998/Math/MathML"> r is the radius of the orbit.

In this scenario, the energy $E$ required to separate a hydrogen atom into a proton and an electron is given as $12.8\text{eV}$, which needs to be converted into joules using the conversion factor $1\text{eV}=1.6\times {10}^{-19}\text{J}$. So,

$12.8\text{eV}=12.8\times 1.6\times {10}^{-19}\text{J}$

We can then substitute the given values into the energy equation and solve for $r$:

$12.8\times 1.6\times {10}^{-19}\text{J}=\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}{2r}$

Solving for $r$, we get:

$r=\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}{2\times 12.8\times 1.6\times {10}^{-19}}$

This simplifies to:

$r=\frac{9\times {10}^{-10}}{16}$

Comparing this with the given form of the radius, which is $\frac{9}{x}\times {10}^{-10}$, we find that the value of$x$ is 16.

The formula to calculate the radius of an orbit for a hydrogen-like atom/ion is:

$r_n=r_0\frac{n^2}{Z}$

where:

• $r_n$ is the radius of the nth orbit,
• $n$ is the principal quantum number (the orbit number),
• $r_0$ is the Bohr radius (radius of the first Bohr orbit in the hydrogen atom), and
• $Z$ is the atomic number (the number of protons in the nucleus).

We're dealing with a Li²⁺ ion and we're interested in the fifth orbit ($n=5$), and given that $r_0$ is 0.51 Å and $Z$ for Li is 3, we can substitute these values into the formula:

$r_5=0.51\times \frac{25}{3}\text{ Å}=4.25\text{ Å}$

which is $4.25\times {10}^{-10}$ m, or equivalently $425\times {10}^{-12}$ m when converted to meters.

The Crrect Answer is

${\mathrm{E}}_H=13.6$

$\begin{array}{rl}& {\mathrm{E}}_{{\mathrm{Li}}^{2+}}=13.6\frac{Z^2}{n^2}=13.6\times \frac{9}{9}=13.6\mathrm{eV}\\ \\ & =136\times {10}^{-1}\mathrm{eV}\end{array}$

The Correct Answer is Option (A)

$\because$ $mvr=\frac{nh}{2\pi }$

$\Rightarrow mv=\frac{nh}{2\pi r}$

The Correct Answer is Option (B)

$\because$ $\overrightarrow{\mu }=\frac{q\overrightarrow{L}}{2m}$

$\Rightarrow \overrightarrow{\mu }=\frac{-e\overrightarrow{L}}{2m}$

The Correct Answer is Option (D)

Radiation is not emitted but absorbed when an electron jumps from low energy to high energy.

Also, E₂ $-$ E₁ is the energy of photon

$\Rightarrow$ E₂ $-$ E₁ = hf

$\Rightarrow f=\frac{E_2-E_1}{h}$

The Correct Answer is Option (D)

We know that $v\propto \frac{Z}{n}$

$\Rightarrow$ Required ratio $=\frac{\frac{2}{3}}{\frac{1}{3}}$

$=2:1$

The Correct Answer is Option (B)

$T.E.=\frac{-Z^{2}m{e}^4}{8{\left(nh{\epsilon }_0\right)}^2}$

$P.E.=\frac{-Z^{2}m{e}^4}{4{\left(nh{\epsilon }_0\right)}^2}$

$K.E.=\frac{Z^{2}m{e}^4}{8{\left(nh{\epsilon }_0\right)}^2}$

As electron makes transition to higher level, total energy and potential energy increases (due to negative sign) while the kinetic energy reduces.

The Correct Answer is Option (C)

An atom based on classical theory of Rutherford's model should collapse as the electrons in continuous circular motion that is a continuously accelerated charge should emit EM waves and so should lose energy. These electrons losing energy should soon fall into heavy nucleus collapsing the whole atom.

The Correct Answer is Option (B)

(A) True, atom of each element emits characteristic spectrum.

(B) True, according to Bohr's postulates $mvr=\frac{nh}{2\pi }$ and hence electron resides into orbits of specific radius called stationary orbits.

(C) False, density of nucleus is constant.

(D) False, A free neutron is unstable decays into proton and electron and antineutrino.

(E) True, unstable nucleus show radioactivity.

The Correct Answer is Option (B)

Resolving power (RP) $\propto$ $\frac{1}{\lambda }$

We know, de-Broglie wavelength

$\lambda =\frac{h}{mv}$

$\therefore$ RP $\propto$ $\frac{mv}{h}$

$\therefore$ $\frac{R{P}_e}{R{P}_p}=\frac{m_e}{m_p}=1837$

The Correct Answer is Option (B)

m_m = 207 m_e

In hydrogen atom one electron present. Now that the electron in hydrogen atom is replaced by a muon ($\mu$).

We know, Energy(E) = $-\frac{e^{4}m}{8{\epsilon }_0^2{n}^2{h}^2}$

For electron,

E_e = $-\frac{e^4{m}_e}{8{\epsilon }_0^2{n}^2{h}^2}$ = 13.6 eV

For muon,

E_m = $-\frac{e^4\left(207\right)m_e}{8{\epsilon }_0^2{n}^2{h}^2}$

= 13.6 $\times$207 = 2815.2 eV

The Correct Answer is Option (A)

We know velocity of electron in n^th shell of hydrogen atom is given by

$v=\frac{2\pi kZ{e}^2}{nh}$

$\therefore$$v\propto \frac{1}{n}$