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The de Broglie wavelength associated with a ball of mass \(1 \mathrm{~kg}\) having a kinetic energy \(0.5 \mathrm{~J}\) is a. \(6.626 \times 10^{-34} \mathrm{~m}\) b. \(13.2 \times 10^{-34} \mathrm{~m}\) c. \(10.38 \times 10^{21} \mathrm{~m}\) d. \(6.626 \AA\)

Short Answer

Expert verified
The correct answer is option (a) \( 6.626 \times 10^{-34} \mathrm{~m} \).

Step by step solution

01

Recall de Broglie Wavelength Formula

The de Broglie wavelength \( \lambda \) of a particle is given by the equation \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \mathrm{~m^2~kg/s} \)) and \( p \) is the momentum of the particle.
02

Calculate Momentum

The momentum \( p \) of an object can be calculated using the relation \( p = \sqrt{2mK} \), where \( m \) is the mass (\( 1 \mathrm{~kg} \)) and \( K \) is the kinetic energy (\( 0.5 \mathrm{~J} \)). Thus, \( p = \sqrt{2 \times 1 \times 0.5} = \sqrt{1} = 1 \mathrm{~kg~m/s} \).
03

Compute de Broglie Wavelength

Substitute the values of \( h \) and \( p \) into the de Broglie equation: \( \lambda = \frac{6.626 \times 10^{-34}}{1} = 6.626 \times 10^{-34} \mathrm{~m} \).
04

Compare and Choose the Correct Answer

The calculated de Broglie wavelength is \( 6.626 \times 10^{-34} \mathrm{~m} \). Compared to the given options, the correct answer is option (a).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. The formula to calculate kinetic energy (KE) for an object with mass (\( m \)) and velocity (\( v \)) is given as:
\[\text{KE} = \frac{1}{2} mv^2\]Understanding kinetic energy is vital, especially when dealing with particles like electrons or even larger objects, because it helps to relate an object's velocity and mass to the energy it contains. It's important to remember:
  • Kinetic energy depends on the square of velocity. This means if you double the velocity, the kinetic energy increases by four times.
  • It's always measured in joules (\( J \)), in the metric system.
In the exercise, the kinetic energy was given as \(0.5 \mathrm{~J}\). Knowing this and the mass of the object, we could link it to momentum, which helps us in further calculations.
Momentum Calculation
Momentum is a key concept in physics. It's used to describe the motion of an object and is a product of its mass and velocity. The formula to calculate momentum (\( p \)) is:
\[p = mv\]In contexts where you have the kinetic energy but want to calculate momentum, there's a useful derived formula:
\[p = \sqrt{2mK}\]This comes handy when directly calculating from kinetic energy. In the problem, given mass (\(1 \mathrm{~kg}\)) and kinetic energy (\(0.5 \mathrm{~J}\)), you calculate momentum as:
\[p = \sqrt{2 \times 1 \times 0.5} = 1 \mathrm{~kg~m/s}\]Understanding momentum's relation to both mass and velocity makes predicting the movement and energy transfer in collisions and other dynamic interactions much easier. It's measured in kilogram meters per second (\(\mathrm{kg~m/s}\)), so watch those units!.
Planck's Constant
Planck's constant is a fundamental constant in nature, denoted as \(h\). It represents the limit below which the laws of classical physics start to change significantly, leading into quantum mechanics. The value of Planck's constant is:
  • \(h = 6.626 \times 10^{-34} \mathrm{~m^2~kg/s}\)
Planck's constant is pivotal in calculations involving quantum mechanics and is essential in the derivation of the de Broglie wavelength.
The de Broglie wavelength (\( \lambda \)) of a particle is determined by \(\lambda = \frac{h}{p}\), connecting the wavelength of a particle to its momentum. This relationship is foundational in wave-particle duality, illustrating that every moving object, including those considered classical like a ball, exhibits wave-like nature.
By using Planck's constant in the exercise, and substituting into the de Broglie equation, we calculated the wavelength of an object as \(6.626 \times 10^{-34} \mathrm{~m}\), showcasing its significance in evaluating quantum scales.

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Most popular questions from this chapter

Which of the following statement(s) is/are correct for orbital angular momentum of \(2 \mathrm{p}\) and 3 p-electron? a. Orbital angular momentum of 3 p-electron is more than that of \(2 p\)-electron b. Orbital angular momentum of 2 p-electron is more than that of \(3 \mathrm{p}\)-electron c. Orbital angular momentum of \(2 \mathrm{p}\) electron is equal to 2 s electron and 3 p electron is equal to 3 s electron d. Orbital angular momentum of 2 p-electron is same as that of 3 p-electron

Hydrogen atoms are excited on \(\mathrm{n}=4\) state. In the spectrum of emitted radiation, number of lines in the ultraviolet and visible regions are respectively a. \(3: 2\) b. \(3: 1\) c. \(1: 3\) d. \(2: 3\)

Same wavelength is observed with respect to \(\mathrm{Li}^{2+}\) from a transition of \(\mathrm{n}_{2}=6\) to \(\mathrm{n}_{1}=3\) a. \(4 \rightarrow 2\) of \(\mathrm{Be}^{+3}\) b. \(3 \rightarrow 1\) of \(\mathrm{H}\)-atom c. \(4 \rightarrow 2\) of \(\mathrm{He}^{+}\) d. \(4 \rightarrow 2\) of He-atom

The ratio of the de-Broglie wavelengths of an \(\alpha\) particle and a proton can be \(2: 1\) when a. Their velocities are in \(4: 1\) ratio b. Their velocities are in \(8: 1\) ratio c. Their kinetic energies are in \(128: 1\) ratio d. Their kinetic energies are in \(256: 1\) ratio.

Which statement is/are true for many electron atoms? a. The \(2 \mathrm{Px}\) and \(2 \mathrm{Py}\) orbitals have the same energy in the absence of an applied magnetic field b. The 2 s and \(2 p\) orbitals are of differing energies, whereas in a hydrogen atom they are same. c. Outer electrons penetrate the electron clouds of inner electrons d. Outer electrons experience the full nuclear charge

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