In hydrogen atom, the electron is moving round the nucleus with velocity $2.18 \times {10^6}\;{\rm{m}}{{\rm{s}}^{ - 1}}$ in an orbit of radius $0.528\;{A^0}$. The acceleration of the electron is

## CORE INFORMATION\n✓ **Answer**: B\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Centripetal Acceleration\n\t**Related Topics**: Circular Motion, Velocity, Radius\n\t**Prerequisites**: Understanding of circular motion and the formula for centripetal acceleration.\n## SOLUTION PATHWAY\n**Given Information**:\n\tVelocity of electron (v) = $2.18 \times 10^6 \; m/s$\n\tRadius of orbit (r) = $0.528 \; A^0 = 0.528 \times 10^{-10} \; m$\n**Method & Steps**:\n\t1. Identify the relevant formula for centripetal acceleration.\n\t\tReasoning: In circular motion, the acceleration is directed towards the center of the circle and is given by $a = \frac{v^2}{r}$.\n\t\tFormula/Rule: $a = \frac{v^2}{r}$\n\t2. Substitute the given values into the formula.\n\t\tWorking: $a = \frac{(2.18 \times 10^6)^2}{0.528 \times 10^{-10}} = \frac{4.7524 \times 10^{12}}{0.528 \times 10^{-10}}$\n\t\tKey Point: Ensure consistent units (meters and seconds).\n\t3. Calculate the acceleration.\n\t\tResult: $a = 9.0 \times 10^{22} \; m/s^2$\n\t\tVerification: The result matches the correct answer option.\n ## OPTION ANALYSIS\n[A] :x: Incorrect calculation of the acceleration.\n[B] ✓: Correct calculation of the centripetal acceleration.\n[C] :x: Incorrect magnitude and sign of the exponent.\n[D] :x: Incorrect magnitude and sign of the exponent.\n ## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1. Incorrect unit conversion: Failing to convert Angstroms to meters.\n\t\t\tHow to avoid: Always ensure all units are in the SI system before calculation.\n\t\t2. Incorrect calculation or exponent handling: Making errors while squaring the velocity or dividing by the radius.\n\t\t\tHow to avoid: Double check numerical calculations and pay attention to the exponents.\n\t**Quick Tips**:\n\t\t1. Remember the centripetal acceleration formula: $a = \frac{v^2}{r}$.\n\t\t2. Use scientific notation while solving to avoid mistakes.\n

$9 \times {10^{18}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$

The correct Option is "$9 \times {10^{22}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$"

$9 \times {10^{ - 22}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$

$9 \times {10^{12}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$

In hydrogen atom, the electron is moving round the nucleus with velocity $2.18 \times {10^6}\;{\rm{m}}{{\rm{s}}^{ - 1}}$ in an orbit of radius $0.528\;{A^0}$. The acceleration of the electron is

## CORE INFORMATION\n✓ **Answer**: B\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Centripetal Acceleration\n\t**Related Topics**: Circular Motion, Velocity, Radius\n\t**Prerequisites**: Understanding of circular motion and the concept of centripetal acceleration.\n## SOLUTION PATHWAY\n**Given Information**:\n\tVelocity (v) = $2.18 \times 10^6 m/s$\n\tRadius (r) = $0.528 A^o = 0.528 \times 10^{-10} m$\n\n**Method & Steps**:\n\t1. **Identify the relevant formula**: The acceleration of an object moving in a circle is given by the centripetal acceleration formula: $a = \frac{v^2}{r}$\n\t\tReasoning: The electron is moving in a circular path around the nucleus, hence we use the centripetal acceleration formula.\n\t\tFormula/Rule: $a = \frac{v^2}{r}$\n\t2. **Substitute the given values**: $a = \frac{(2.18 \times 10^6 m/s)^2}{0.528 \times 10^{-10} m}$\n\t\tWorking: Calculate the square of velocity and substitute the radius in meters.\n\t\tKey Point: Ensure consistent units (meters and seconds).\n\t3. **Calculate the acceleration**: $a = \frac{4.7524 \times 10^{12}}{0.528 \times 10^{-10}} = 8.999 \times 10^{22} \approx 9 \times 10^{22} m/s^2$\n\t\tResult: The acceleration is approximately $9 \times 10^{22} m/s^2$\n\t\tVerification: The calculated value is consistent with the correct answer.\n## OPTION ANALYSIS\n[A] :x: Incorrect calculation.\n[B] ✓: Correct application of centripetal acceleration formula and calculation.\n[C] :x: Incorrect calculation. The exponent is wrong.\n[D] :x: Incorrect calculation. The exponent is wrong.\n## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1. **Unit Conversion**: Incorrect conversion of Angstroms to meters can lead to wrong results. Remember $1 A^o = 10^{-10} m$.\n\t\t2. **Formula Application**: Confusing centripetal acceleration formula with other formulas. Ensure you use $a = \frac{v^2}{r}$.\n\t**Quick Tips**:\n\t\t1. **Approximation**: In multiple-choice questions, you can approximate the values to get a quicker answer. In this case, $2.18 \approx 2$ and $0.528 \approx 0.5$.\n\t\t2. **Exponent Handling**: Pay close attention to exponents while doing calculations to avoid errors.

$9 \times {10^{18}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$

The correct Option is "$9 \times {10^{22}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$"

$9 \times {10^{ - 22}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$

$9 \times {10^{12}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$

In hydrogen atom, the electron is moving round the nucleus with velocity $2.18 \times {10^6}\;{\rm{m}}{{\rm{s}}^{ - 1}}$ in an orbit of radius $0.528\;{A^0}$. The acceleration of the electron is

## CORE INFORMATION\n✓ **Answer**: B\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Centripetal Acceleration\n\t**Related Topics**: Circular Motion, Velocity, Radius\n\t**Prerequisites**: Understanding of circular motion and the formula for centripetal acceleration.\n## SOLUTION PATHWAY\n**Given Information**:\n\tVelocity (v) = $2.18 \times 10^6 m/s$\n\tRadius (r) = $0.528 A^0 = 0.528 \times 10^{-10} m$\n**Method & Steps**:\n\t1. Identify the appropriate formula for centripetal acceleration.\n\t\tReasoning: Since the electron is moving in a circular orbit, it experiences centripetal acceleration.\n\t\tFormula/Rule: $a_c = \frac{v^2}{r}$\n\t2. Substitute the given values into the formula.\n\t\tWorking: $a_c = \frac{(2.18 \times 10^6)^2}{0.528 \times 10^{-10}}$\n\t\tKey Point: Ensure consistent units (meters and seconds).\n\t3. Calculate the result.\n\t\tResult: $a_c = \frac{4.7524 \times 10^{12}}{0.528 \times 10^{-10}} = 8.99 \times 10^{22} m/s^2 \approx 9 \times 10^{22} m/s^2$\n\t\tVerification: The calculated value matches the correct option.\n ## OPTION ANALYSIS\n[A] :x: Incorrect. This option has an exponent that is too low.\n[B] ✓: Correct. The calculated centripetal acceleration is approximately $9 \times 10^{22} m/s^2$.\n[C] :x: Incorrect. This option has an incorrect sign and exponent.\n[D] :x: Incorrect. This option has an exponent that is too low.\n ## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1. Unit Conversion: Not converting angstroms to meters can lead to a wrong answer.\n\t\t2. Calculation Errors: Mistakes in squaring the velocity or in dividing can also lead to incorrect results.\n\t**Quick Tips**:\n\t\t1. Remember the formula for centripetal acceleration ($a_c = v^2/r$).\n\t\t2. Always check the units before calculations.

$9 \times {10^{18}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$

The correct Option is "$9 \times {10^{22}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$"

$9 \times {10^{ - 22}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$

$9 \times {10^{12}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$

In hydrogen atom, the electron is moving round the nucleus with velocity $2.18 \times {10^6}\;{\rm{m}}{{\rm{s}}^{ - 1}}$ in an orbit of radius $0.528\;{A^0}$. The acceleration of the electron is

## CORE INFORMATION\n✓ **Answer**: B\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Centripetal Acceleration\n\t**Related Topics**: Circular Motion, Velocity, Radius\n\t**Prerequisites**: Understanding of circular motion and the formula for centripetal acceleration.\n## SOLUTION PATHWAY\n**Given Information**:\n\tVelocity (v) = $2.18 \times 10^6 m/s$\n\tRadius (r) = $0.528 \overset{\circ}{A} = 0.528 \times 10^{-10} m$\n\t**Important conditions**: The electron is moving in a circular path around the nucleus.\n**Method & Steps**:\n\t1.\tIdentify the relevant formula.\n\t\tReasoning: In circular motion, the acceleration towards the center of circle is centripetal acceleration.\n\t\tFormula/Rule: $a = v^2/r$\n\t2.\tSubstitute the given values into the formula.\n\t\tWorking: $a = (2.18 \times 10^6 m/s)^2 / (0.528 \times 10^{-10} m)$\n\t\tKey Point: Ensure consistent units.\n\t3.\tCalculate the acceleration.\n\t\tResult: $a = (4.7524 \times 10^{12}) / (0.528 \times 10^{-10}) = 8.99 \times 10^{22} m/s^2 \approx 9 \times 10^{22} m/s^2$\n\t\tVerification: The calculated answer matches with correct answer.\n## OPTION ANALYSIS\n[A] :x: Incorrect calculation.\n[B] ✓: Correctly calculated using the formula for centripetal acceleration.\n[C] :x: Incorrect calculation.\n[D] :x: Incorrect calculation.\n## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1.[Error 1]: Forgetting to convert the radius from angstroms to meters. [How to avoid]: Always convert all units to the standard SI units before calculation.\n\t\t2.[Error 2]: Incorrectly applying the centripetal acceleration formula. [How to avoid]: Recall the correct formula ($a = v^2/r$) and ensure the values are put in their correct places.\n\t**Quick Tips**:\n\t\t1.[Helpful shortcut]: Remember that centripetal acceleration is always directed towards the center of the circle.\n\t\t2.[Memory aid]: The formula $a=v^2/r$ can be remembered as 'velocity squared over radius'.

$9 \times {10^{18}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$

The correct Option is "$9 \times {10^{22}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$"

$9 \times {10^{ - 22}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$

$9 \times {10^{12}}{\rm{\;m}}{{\rm{s}}^{ - 2}}$

NEET Physics Nuclei MCQs

Prepare for NEET Physics Nuclei with MCQs & PYQs on NEET.GUIDE. Enjoy free practice, previous year questions, and expert explanations to understand nuclear structure and reactions.

NEET Questions / Physics / Nuclei

Select Topic

🎓 Ncert Level:Application

🎯Difficulty :Medium

⚡ Time Required :2 mins

🔄Ncert Concepts :Circular Motion, Centripetal Acceleration

10.

In hydrogen atom, the electron is moving round the nucleus with velocity $2.18 imes {10^6}\;{\rm{m}}{{\rm{s}}^{ - 1}}$ in an orbit of radius $0.528\;{A^0}$ . The acceleration of the electron is