Q.An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy $10\ \text{keV}$, and the second with $100\ \text{keV}$. Which is faster, the electron or the proton? Obtain the ratio of their speeds. (electron mass $= 9.11 \times 10^{-31}\ \text{kg}$, proton mass $= 1.67 \times 10^{-27}\ \text{kg}$, $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$).

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.The magnetic field required to accelerate deuterons in a cyclotron operated at a frequency of 28 MHz is (Mass of proton = $1.67 \times 10^{-27}$ kg) (A) 7.348 T (B) 0.917 T (C) 1.837 T (D) 3.674 T

   ›Reveal solutionSolution

   The cyclotron resonance condition $f = \dfrac{qB}{2\pi m}$ gives $B = \dfrac{2\pi m f}{q} = 3.674~\text{T}$ for a deuteron.

   Given

   • Operating frequency $f = 28~\text{MHz} = 28 \times 10^{6}~\text{Hz}$
   • Deuteron: charge $q = 1.6 \times 10^{-19}~\text{C}$, mass $m = 2 m_p = 2(1.67 \times 10^{-27}) = 3.34 \times 10^{-27}~\text{kg}$

   Cyclotron condition

   The accelerating frequency must equal the cyclotron frequency:

   $f = \frac{qB}{2\pi m} \quad\Rightarrow\quad B = \frac{2\pi m f}{q}$ …

   🔒 Start your 14-day free trial to see the full solution →
2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.In $\beta$-decay, when a proton converts into a neutron, the particles emitted are (A) Electron, neutrino (B) Positron, neutrino (C) Positron, antineutrino (D) Electron, antineutrino

   ›Reveal solutionSolution

   In $\beta$-decay, a proton converts into a neutron only in positron emission (a type of $\beta^+$ decay), which emits a positron and a neutrino. The correct option is (B).

   The key to this question is understanding that a proton is lighter than a neutron. For a proton to turn into a neutron, the process must consume energy — it cannot happen spontaneously inside a nucleus unless that energy is supplied by the nucleus itself. This is the opposite of neutron-to-proton conversion, which releases energy.

   Let’s break down the two main types of beta decay:

   1. $\beta^-$ decay (neutron → proton): A neutron inside the nucleus decays into a proton, an electron, and an antineutrino.

   $n \rightarrow p + e^- + \bar{\nu}_e$

   Here, the emitted particles are an electron and an antineutrino. This is the common beta decay of neutron-rich nuclei.

   2. $\beta^+$ decay (proton → neutron): A proton inside the nucleus converts into a neutron, a positron, and a neutrino.

   $p \rightarrow n + e^+ + \nu_e$

   This can only happen if the nucleus has enough extra energy (from its internal binding) to make up for the mass difference — the neutron is heavier than the proton by about $1.3 \text{ MeV}/c^2$. The emitted particles are a positron and a neutrino. …

   🔒 Start your 14-day free trial to see the full solution →
3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.A ball P in motion collides with another ball Q of same mass at rest. If the coefficient of restitution between the two balls is 0.2, then the ratio of the velocities of the two balls P and Q after collision is (A) $1 : 3$ (B) $3 : 4$ (C) $1 : 1$ (D) $2 : 3$

   ›Reveal solutionSolution

   Momentum conservation plus the restitution equation for equal masses give $v_P=0.4u$ and $v_Q=0.6u$, so $v_P:v_Q = 2:3$ — option (D).

   Let each mass be $m$, P's initial speed be $u$, and Q be at rest. After the head-on collision let P have speed $v_1$ and Q have speed $v_2$.

   Conservation of momentum.

   $mu = mv_1+mv_2 \;\Rightarrow\; v_1+v_2 = u.$

   Coefficient of restitution.

   $e = \frac{v_2-v_1}{u} = 0.2 \;\Rightarrow\; v_2-v_1 = 0.2u.$

   Solve the pair. Adding and subtracting: …

   🔒 Start your 14-day free trial to see the full solution →
4. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $e$, $m$ and $\tau$ are charge, mass and average collision time for an electron respectively, then the magnitude of drift velocity per unit electric field is (A) $\dfrac{em}{\tau}$ (B) $\dfrac{e\tau}{m}$ (C) $\dfrac{m\tau}{e}$ (D) $\dfrac{\tau}{me}$

   ›Reveal solutionSolution

   The drift velocity per unit electric field is the electron mobility, derived from balancing electric and damping forces; the result is $\frac{e\tau}{m}$, which corresponds to option (B).

   The key concept here is electron mobility — how fast an electron drifts in response to an applied electric field, given that it constantly collides with the lattice. The drift velocity isn't constant acceleration; instead, electrons accelerate between collisions, then get randomized. The average effect gives a steady drift speed proportional to the field. The proportionality constant is exactly what the question asks for: drift velocity per unit electric field.

   1. Start with the equation of motion for an electron in a field An electron of mass $m$ and charge $e$ experiences a force $F = eE$ from an electric field $E$. Between collisions, it accelerates. But collisions with ions/impurities act like a damping force, leading to a steady average velocity. The standard model gives:

   $m \frac{dv_d}{dt} = eE - \frac{m v_d}{\tau}$

   where $v_d$ is the drift velocity and $\tau$ is the average time between collisions (relaxation time).

   2. Steady-state condition In steady state, the drift velocity is constant, so $\frac{dv_d}{dt} = 0$. Then:

   $0 = eE - \frac{m v_d}{\tau}$

   Rearranging:

   $\frac{m v_d}{\tau} = eE$

   $v_d = \frac{e\tau}{m} E$

   3. Extract the required quantity The question asks for the magnitude of drift velocity per unit electric field, i.e., $\frac{v_d}{E}$. From the equation above: $$$$

   🔒 Start your 14-day free trial to see the full solution →
5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.An electron and a positron enter a uniform electric field $E$ perpendicular to it with equal speeds at the same time. The distance of separation between them in the direction of the field after a time ‘t’ is $\left(\dfrac{e}{m}\ \text{is specific charge of electron}\right)$ (A) $\dfrac{2Eet^2}{m}$ (B) $\dfrac{Eet^2}{m}$ (C) $\dfrac{Eet^2}{2m}$ (D) Zero

   ›Reveal solutionSolution

   The key idea is that the electron and positron have opposite charges, so they accelerate in opposite directions in the electric field. Their separation in the field direction is twice the distance each travels, giving $\frac{2Eet^2}{m}$.

   Concept & Intuition

   An electric field exerts a force on a charged particle: $F = qE$. Since the electron has charge $-e$ and the positron has charge $+e$, they experience forces in opposite directions. Both start with the same initial speed perpendicular to the field, so their motion along the field is purely due to this force. The distance each moves along the field is given by kinematics for constant acceleration. Their separation is the sum of the distances they move apart.

   Step-by-step solution

   1. Determine the acceleration of each particle

      The force on a charge $q$ in a uniform electric field $E$ is $F = qE$. By Newton’s second law, $F = ma$, so the acceleration is $a = \frac{qE}{m}$.

      For the electron: $q = -e$, so $a_e = -\frac{eE}{m}$ (negative direction).

      For the positron: $q = +e$, so $a_p = +\frac{eE}{m}$ (positive direction).

      The magnitude of acceleration for both is $a = \frac{eE}{m}$.

   2. Initial conditions along the field

      The particles enter with equal speeds perpendicular to the field. That means their initial velocity along the field is zero. So along the field direction, each starts from rest.

   3. Distance traveled along the field in time $t$

      Using $s = ut + \frac{1}{2} a t^2$ with $u = 0$, the distance each moves from the starting point is

   $s = \frac{1}{2} a t^2 = \frac{1}{2} \left( \frac{eE}{m} \right) t^2.$

   The electron moves a distance $s$ in the negative direction, the positron moves $s$ in the positive direction.

   4. Separation between them The total separation along the field direction is the distance from the electron’s position to the positron’s position. Since they move in opposite directions, the separation is …

   🔒 Start your 14-day free trial to see the full solution →
6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Two radioactive substances A and B have same number of initial nuclei. If the half-lives of A and B are 1.5 days and 4.5 days respectively, then the ratio of the number of nuclei remaining in A and B after 9 days is (A) 1 : 16 (B) 1 : 1 (C) 1 : 4 (D) 1 : 8

   ›Reveal solutionSolution

   The key is to use the exponential decay law $N = N_0 (1/2)^{t/T}$. After 9 days, A has undergone 6 half-lives and B has undergone 2 half-lives, so the ratio $N_A : N_B = (1/2)^6 : (1/2)^2 = 1/64 : 1/4 = 1 : 16$. The correct option is (A).

   Concept and Intuition

   Radioactive decay follows a simple rule: after each half-life, the number of remaining nuclei is halved. The half-life is the time it takes for half of a sample to decay. So if you know how many half-lives have passed, you can directly compute the fraction left as $(1/2)^{\text{number of half-lives}}$. This avoids messy exponential formulas — just count the half-lives.

   Step-by-step solution

   1. Identify the initial condition

      Both substances A and B start with the same number of nuclei, call it $N_0$. So $N_{A0} = N_{B0} = N_0$.

   2. Determine the number of half-lives elapsed for each

      • For A: half-life $T_A = 1.5$ days, time $t = 9$ days. Number of half-lives: $n_A = \frac{t}{T_A} = \frac{9}{1.5} = 6$.
      • For B: half-life $T_B = 4.5$ days. Number of half-lives: $n_B = \frac{9}{4.5} = 2$.

   3. Apply the decay law

      After $n$ half-lives, the remaining number is $N = N_0 \left(\frac{1}{2}\right)^n$.

      • For A: $N_A = N_0 \left(\frac{1}{2}\right)^6 = N_0 \cdot \frac{1}{64}$.
      • For B: $N_B = N_0 \left(\frac{1}{2}\right)^2 = N_0 \cdot \frac{1}{4}$.

   4. Compute the ratio

      $$$$

   🔒 Start your 14-day free trial to see the full solution →
7. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.The distance of closest approach of an alpha particle from a nucleus when the alpha particle moves towards a nucleus with a kinetic energy ‘E’ is ‘x’. The distance of closest approach when the alpha particle approaches the same nucleus with kinetic energy $0.4E$ is (A) $3.5x$ (B) $5x$ (C) $2.5x$ (D) $4x$

   ›Reveal solutionSolution

   The distance of closest approach is inversely proportional to the kinetic energy of the alpha particle. When the kinetic energy is reduced to $0.4E$, the distance increases by a factor of $2.5$, so the answer is $2.5x$.

   The distance of closest approach in Rutherford scattering is the point where the alpha particle’s kinetic energy is completely converted into electrostatic potential energy due to repulsion from the nucleus. At that instant, the particle stops momentarily before turning back.

   The potential energy at a distance $r$ from the nucleus (charge $Ze$) is given by Coulomb’s law:

   $U = \frac{1}{4\pi\epsilon_0} \cdot \frac{(2e)(Ze)}{r}$

   where $2e$ is the charge of the alpha particle. At the distance of closest approach $r_0$, all kinetic energy $E$ becomes this potential energy:

   $E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2Ze^2}{r_0}$

   This shows that $E$ is inversely proportional to $r_0$ — a larger kinetic energy lets the alpha particle get closer before being repelled, while a smaller kinetic energy stops it farther away.

   1. For the first case, kinetic energy $E$ gives distance $x$:

   $E = \frac{k}{x} \quad \text{where } k = \frac{2Ze^2}{4\pi\epsilon_0}$

   2. For the second case, kinetic energy $0.4E$ gives a new distance $r$:

   $0.4E = \frac{k}{r}$

   3. Divide the second equation by the first: …

   🔒 Start your 14-day free trial to see the full solution →
8. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.The half-life period of element X is same as the mean life time of element Y. Assume initially X and Y have same number of atoms. Then (A) Initially X and Y have same decay rates (B) Always X and Y decay at same rate (C) Y decays faster than X (D) X decays faster than Y

   ›Reveal solutionSolution

   The half-life of X equals the mean life of Y, so the decay constant of X is smaller than that of Y; thus Y decays faster. The correct option is (C).

   Concept & Intuition

   The key is understanding the relationship between half-life ($t_{1/2}$), mean life ($\tau$), and the decay constant ($\lambda$). For any radioactive decay,

   $t_{1/2} = \frac{\ln 2}{\lambda}, \quad \tau = \frac{1}{\lambda}.$

   If the half-life of X equals the mean life of Y, then

   $\frac{\ln 2}{\lambda_X} = \frac{1}{\lambda_Y}.$

   This implies $\lambda_Y = \frac{\lambda_X}{\ln 2}$. Since $\ln 2 \approx 0.693 < 1$, we get $\lambda_Y > \lambda_X$. A larger decay constant means faster decay — more decays per unit time for the same number of atoms.

   Step-by-step reasoning

   1. Write the given condition mathematically

      Half-life of X: $t_{1/2}^{(X)} = \frac{\ln 2}{\lambda_X}$.

      Mean life of Y: $\tau_Y = \frac{1}{\lambda_Y}$.

      Given: $t_{1/2}^{(X)} = \tau_Y$ → $\frac{\ln 2}{\lambda_X} = \frac{1}{\lambda_Y}$.

   2. Solve for the ratio of decay constants

      Rearranging: $\lambda_Y = \frac{\lambda_X}{\ln 2}$.

      Since $\ln 2 < 1$, $\lambda_Y > \lambda_X$.

   3. Interpret the decay rate

      Decay rate (activity) $R = \lambda N$. Initially, $N_X = N_Y$.

      So $R_X = \lambda_X N$, $R_Y = \lambda_Y N$.

      Because $\lambda_Y > \lambda_X$, we have $R_Y > R_X$ initially.

      Thus Y decays faster than X at the start.

   4. Check if the rates ever become equal

      As time passes, the number of atoms changes: $N_X(t) = N_0 e^{-\lambda_X t}$, $N_Y(t) = N_0 e^{-\lambda_Y t}$.

      The ratio of activities: $\frac{R_Y(t)}{R_X(t)} = \frac{\lambda_Y e^{-\lambda_Y t}}{\lambda_X e^{-\lambda_X t}} = \frac{\lambda_Y}{\lambda_X} e^{-(\lambda_Y - \lambda_X)t}$. …

   🔒 Start your 14-day free trial to see the full solution →
9. TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
   Q.Two objects I and II of mass $0.8\, \text{kg}$ and $0.6\, \text{kg}$ collide elastically. If before collision, the object II was at rest and object I was moving. Find the ratio of the final velocity of object I and object II. (A) $6 : 7$ (B) $4 : 3$ (C) $1 : 8$ (D) $1 : 16$

   ›Reveal solutionSolution

   For an elastic collision where one object is initially at rest, the final velocities are given by formulas derived from conservation of momentum and kinetic energy. Here, with masses $0.8\,\text{kg}$ and $0.6\,\text{kg}$, the ratio of final velocities is $1:8$, so option (C) is correct.

   ---

   Concept and Intuition

   In any collision, momentum is always conserved. In an elastic collision, kinetic energy is also conserved. When one object is initially at rest, these two conservation laws are enough to solve for the final velocities in terms of the masses and the initial velocity. The key insight: the formulas that emerge are simple and depend only on the mass ratio. No need to know the actual initial speed — the ratio of final velocities is fixed by the masses alone.

   ---

   Step-by-Step Solution

   1. Set up the problem with variables.

   Let:

   • $m_1 = 0.8\,\text{kg}$ (object I, initially moving)
   • $m_2 = 0.6\,\text{kg}$ (object II, initially at rest)
   • $u_1 = u$ (initial velocity of I, unknown but not needed)
   • $u_2 = 0$
   • $v_1$ = final velocity of I
   • $v_2$ = final velocity of II

   We want the ratio $v_1 : v_2$.

   2. Apply conservation of momentum.

   Total momentum before = total momentum after:

   $m_1 u + m_2 \cdot 0 = m_1 v_1 + m_2 v_2$

   $0.8 u = 0.8 v_1 + 0.6 v_2 \quad \text{(Equation 1)}$

   3. Apply conservation of kinetic energy (elastic collision).

   Total kinetic energy before = total kinetic energy after:

   $\frac{1}{2} m_1 u^2 + 0 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$

   $0.8 u^2 = 0.8 v_1^2 + 0.6 v_2^2 \quad \text{(Equation 2)}$

   4. Use the standard elastic collision result for a stationary target.

   For an elastic collision where $u_2 = 0$, there are well-known formulas:

   $v_1 = \frac{m_1 - m_2}{m_1 + m_2} \, u$

   $v_2 = \frac{2 m_1}{m_1 + m_2} \, u$

   These come from solving the two equations above. Let’s verify quickly:

   ›Proof

   Derivation of the formulas

   From momentum: $m_1 u = m_1 v_1 + m_2 v_2$ → $v_2 = \frac{m_1}{m_2}(u - v_1)$.

   From energy: $m_1 u^2 = m_1 v_1^2 + m_2 v_2^2$. Substitute $v_2$:

   $m_1 u^2 = m_1 v_1^2 + m_2 \left[\frac{m_1}{m_2}(u - v_1)\right]^2$

   Simplify: $m_1 u^2 = m_1 v_1^2 + \frac{m_1^2}{m_2}(u - v_1)^2$

   Divide by $m_1$: $u^2 = v_1^2 + \frac{m_1}{m_2}(u - v_1)^2$

   Multiply by $m_2$: $m_2 u^2 = m_2 v_1^2 + m_1(u^2 - 2u v_1 + v_1^2)$

   Rearranging: $0 = (m_1 + m_2) v_1^2 - 2 m_1 u v_1 + (m_1 - m_2) u^2$ …

   🔒 Start your 14-day free trial to see the full solution →
10. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.A moving particle collides with a stationary particle of mass $\dfrac{1}{n}$ times the mass of moving particle, the fraction of its kinetic energy transferred to the stationary particle is (A) $\dfrac{4n^2}{(1+n)^2}$ (B) $\dfrac{4n}{(1+n)^2}$ (C) $\dfrac{4n}{1+n^2}$ (D) $4n^2$

    ›Reveal solutionSolution

    In a one-dimensional elastic collision, the fraction of kinetic energy transferred from the moving particle to the stationary one depends only on the mass ratio. The answer is $\dfrac{4n}{(1+n)^2}$.

    The problem is about energy transfer in a collision. When a moving particle hits a stationary one, the amount of kinetic energy that gets passed along depends on how their masses compare. For an elastic collision (kinetic energy is conserved overall), the fraction transferred to the stationary particle can be found using the velocities after impact.

    The key is to realize that the fraction of kinetic energy transferred is not the same as the fraction of momentum transferred. Energy depends on the square of velocity, so we need the final speed of the stationary particle. The standard result for a one-dimensional elastic collision gives that speed directly in terms of the masses.

    Let the moving particle have mass $m$ and initial velocity $u$. The stationary particle has mass $m/n$. After the collision, let their velocities be $v_1$ and $v_2$ respectively.

    1. Apply conservation of momentum. Initial momentum: $m u + 0 = m u$. Final momentum: $m v_1 + (m/n) v_2$. So:

    $m u = m v_1 + \frac{m}{n} v_2 \quad \Rightarrow \quad u = v_1 + \frac{v_2}{n}.$

    2. Apply conservation of kinetic energy (elastic collision). Initial KE: $\frac12 m u^2$. Final KE: $\frac12 m v_1^2 + \frac12 \left(\frac{m}{n}\right) v_2^2$. So:

    $\frac12 m u^2 = \frac12 m v_1^2 + \frac12 \frac{m}{n} v_2^2 \quad \Rightarrow \quad u^2 = v_1^2 + \frac{v_2^2}{n}.$

    3. Solve for $v_2$. From the momentum equation, $v_1 = u - \frac{v_2}{n}$. Substitute into the energy equation:

    $u^2 = \left(u - \frac{v_2}{n}\right)^2 + \frac{v_2^2}{n}.$

    Expand:

    $u^2 = u^2 - \frac{2u v_2}{n} + \frac{v_2^2}{n^2} + \frac{v_2^2}{n}.$

    Cancel $u^2$ from both sides:

    $0 = -\frac{2u v_2}{n} + v_2^2 \left(\frac{1}{n^2} + \frac{1}{n}\right).$

    Factor $v_2$ (non-zero after collision):

    $\frac{2u}{n} = v_2 \left(\frac{1}{n^2} + \frac{1}{n}\right).$

    Simplify the bracket: $\frac{1}{n^2} + \frac{1}{n} = \frac{1 + n}{n^2}$.

    So:

    $v_2 = \frac{2u}{n} \cdot \frac{n^2}{1+n} = \frac{2u n}{1+n}.$

    4. Find the kinetic energy transferred. Initial KE of moving particle: $K_i = \frac12 m u^2$. …

    🔒 Start your 14-day free trial to see the full solution →
11. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.A large metal plate has a surface charge density of $8.85 \times 10^{-6}$ C/m$^2$. An electron having initial kinetic energy of $8 \times 10^{-17}$ J is moving towards the center of the plate. If the electron stops just before reaching the plate then the initial distance between the electron and the plate is [Take $\epsilon_0 = 8.85 \times 10^{-12}$ C$^2$/Nm$^2$] (A) 0.5 mm (B) 0.1 mm (C) 0.2 cm (D) 0.02 cm

    ›Reveal solutionSolution

    The field just outside the charged metal plate is $E=\sigma/\epsilon_0=10^{6}\ \text{V/m}$. Equating the work done against this field to the electron's kinetic energy, $d=\dfrac{KE}{eE}=0.5\ \text{mm}$ — option (A).

    For a large (conducting) metal plate carrying surface charge density $\sigma$, the electric field just outside its surface is

    $E=\frac{\sigma}{\epsilon_0}=\frac{8.85\times10^{-6}}{8.85\times10^{-12}}=1.0\times10^{6}\ \text{V/m}.$

    The electron is stopped by the retarding force $F=eE$ over the distance $d$. All its kinetic energy is spent as work done against the field:

    $KE=eE\,d\ \Rightarrow\ d=\frac{KE}{eE}.$ …

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
    Q.Find the mobility of electron in a wire, if its average collision time is $9.1 \times 10^{-15}$ s. (charge of electron $= 1.6 \times 10^{-19}$ C and mass of electron $= 9.1 \times 10^{-31}$ kg) (A) $9.1 \times 10^{-3}$ m$^2$/V-s (B) $1.6 \times 10^{-3}$ m$^2$/V-s (C) $1.75 \times 10^{-3}$ m$^2$/V-s (D) $1 \times 10^{-3}$ m$^2$/V-s

    ›Reveal solutionSolution

    Mobility is the drift velocity per unit electric field, and for electrons it equals $\mu = \frac{e\tau}{m}$. Substituting the given values gives $\mu = 1.6 \times 10^{-3} \ \text{m}^2/\text{V-s}$, which matches option (B).

    The concept here is mobility — a measure of how quickly a charge carrier can move through a conductor when an electric field is applied. For electrons in a wire, mobility depends directly on the average time between collisions (the relaxation time $\tau$) and inversely on the electron's mass. The charge $e$ appears in the numerator because a larger charge means a stronger force from the same field.

    The formula $\mu = \frac{e\tau}{m}$ comes from equating the electric force $eE$ to the drag force $m v_d / \tau$ (from collisions) in the steady state, giving drift velocity $v_d = \frac{eE\tau}{m}$, and then $\mu = v_d/E$.

    Let’s work through the numbers.

    1. Write the formula Mobility of an electron:

    $\mu = \frac{e \tau}{m}$

    2. Substitute the given values $e = 1.6 \times 10^{-19} \ \text{C}$, $\tau = 9.1 \times 10^{-15} \ \text{s}$, $m = 9.1 \times 10^{-31} \ \text{kg}$

    $\mu = \frac{(1.6 \times 10^{-19})(9.1 \times 10^{-15})}{9.1 \times 10^{-31}}$

    3. Simplify step by step The $9.1$ in numerator and denominator cancel: …

    🔒 Start your 14-day free trial to see the full solution →