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Wave nature of matter is not expericnced in our daily life because the value of wavelength is (1) very large (2) very small (3) lies in the ultraviolet region (4) lies in the infrared region

Short Answer

Expert verified
The wavelength is very small. Hence, wave nature of matter is not experienced in our daily life.

Step by step solution

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01

Understanding the Concept

Matter exhibits wave-like properties, described by the de Broglie wavelength equation \[ \lambda = \frac{h}{mv} \] where \( \lambda \) is the wavelength, \( h \) is Planck's constant, \( m \) is the mass, and \( v \) is the velocity of the particle.
02

Apply the de Broglie Equation

For everyday objects (with large mass and moderate velocity), the de Broglie wavelength \( \lambda \) is extremely small because the denominator \( mv \) is very large.
03

Evaluate the Wavelength Magnitude

Calculate \( \lambda = \frac{h}{mv} \) using approximate values for macroscopic objects. For example, with \( m \) in kilograms and \( v \) in meters per second, \( \lambda \) turns out to be on the order of \( 10^{-35} \) meters, which is extremely small.
04

Conclusion Based on Evaluation

Since the de Broglie wavelength for macroscopic objects is extremely small, it is not observed in our daily life. Therefore, the correct answer is option (2) very small.

Key Concepts

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

de Broglie wavelength
The concept of de Broglie wavelength is central to understanding the wave nature of matter. Introduced by Louis de Broglie in 1924, it suggests that every particle has a wavelength associated with it. The equation used to calculate this wavelength is:
\[ \( \lambda = \frac{h}{mv} \) \]
Here, \( \lambda \) represents the de Broglie wavelength, \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity. This equation shows that the wavelength of a particle is inversely proportional to its momentum (product of mass and velocity).
The de Broglie wavelength becomes significant in the quantum realm, where particles have very small masses. This means that for electrons, atoms, and other microscopic particles, the wavelength is large enough to measure and observe. But for macroscopic objects like humans or cars, the wavelength is extremely small, making it almost impossible to detect.
Planck's constant
Planck's constant \(\, h \,\) is a fundamental constant in physics, introduced by Max Planck. It plays a critical role in quantum mechanics and is essential for the de Broglie equation. Its value is approximately \(\, 6.626 \times 10^{-34} \, \) joule-seconds.
Planck's constant signifies the scale at which quantum mechanical effects become noticeable. It provides a bridge between the macroscopic and microscopic worlds. In the de Broglie wavelength formula, \(\, h \,\) represents the proportionality factor that links the wave characteristics of a particle to its momentum.
In simple terms, Planck's constant is crucial for calculating the wavelength of particles. Its extremely small value also explains why the wave nature of large objects isn't noticeable. The very tiny wavelength calculated using macroscopic masses and velocities demonstrates why everyday objects don’t exhibit observable wave-like properties.
macroscopic objects
Macroscopic objects refer to items that can be seen with the naked eye, such as a baseball or a car. When examining the wave nature of these objects using the de Broglie wavelength, the results are fascinating.
Let's apply the de Broglie equation \(\, \lambda = \frac{h}{mv} \,\) to a common object, say a ball weighing 0.145 kg (about 5 ounces) moving at a speed of 30 m/s (approximately 108 km/h). Plugging these values into the equation, the wavelength \(\, \lambda \,\) would be:
\[ \lambda = \frac{6.626 \times 10^{-34}}{0.145 \times 30} \approx 1.52 \times 10^{-34} \,\text{meters} \]
This result shows an insignificantly small wavelength. That's why we don't observe wave-like behavior in our everyday lives. The immense mass and typical speeds of macroscopic objects render their de Broglie wavelengths too tiny to detect or influence their observable properties. This is fundamentally why the wave nature of matter is not experienced in our day-to-day interactions.

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Most popular questions from this chapter

Which of the following statements is wrong? (1) \(\alpha\) -rays, \(\beta\) -rays and cathode rays consist of particles of matter. (2) The addition of a neutron to the nucleus of an atom do not affects its chemical propertics. (3) When the speed of electron increases its specific charge decreases. (4) The fundamental particle with highest specific charge is proton.

Which do not explains correctly about the \(e / m\) (specific charge)? (1) The \(\mathrm{e} / \mathrm{m}\) is not constant for positive rays. (2) The ratio of the \(\mathrm{e} / \mathrm{m}\) of an electron to that of a hydrogen ion is \(1840: 1\). (3) If \(\mathrm{S}_{1}\) is the \(e / m\) of cathode rays and \(\mathrm{S}_{2}\) is the \(\mathrm{e} / \mathrm{m}\) of positive rays then \(\mathrm{S}_{1}>\mathrm{S}_{2}\) (4) The specific charge of positive rays is much less than the specific charge for cathode rays because charge in positive rays is less.

To move an electron in a hydrogen atom from the ground state to the second excited state, \(12.084 \mathrm{eV}\) is required. IIow much energy is required to cause one mole of hydrogen atoms to undergo this transition? (1) \(984 \mathrm{~kJ}\) (2) \(1036 \mathrm{~kJ}\) (3) \(1166 \mathrm{~kJ}\) (4) \(1312 \mathrm{~kJ}\)

The region of space where there is maximum probability of finding an electron at any instant is (1) an orbit (2) an orbital (3) a stationary state (4) subshell

Which electronic configuration does not follow Pauli's exclusion principle? (1) \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{4}\) (2) \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{4} 4 \mathrm{~s}^{2}\) (3) \(1 s^{2} 2 p^{4}\) (4) \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{3}\)

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