Wave Nature of Matter

A proton and an $\alpha$-particle are accelerated from rest by $2 \mathrm{V}$ and $4 \mathrm{V}$ potentials, respectively. The ratio of their de-Broglie wavelength is :

The de Broglie wavelength of an electron having kinetic energy $E$ is $\lambda$. If the kinetic energy of electron becomes $\frac{E}{4}$, then its de-Broglie wavelength will be:

The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy:
(Assume $m_p=m_e\times 1849$)

Proton $\left(\mathrm{P}\right)$ and electron $\left(\mathrm{e}\right)$ will have same de-Broglie wavelength when the ratio of their momentum is (assume, $m_p=1849$ $m_e$)

The de Broglie wavelength of a molecule in a gas at room temperature $\left(300 \mathrm{K}\right)$ is ${\lambda }_1$. If the temperature of the gas is increased to $600 \mathrm{K}$, then the de Broglie wavelength of the same gas molecule becomes

The kinetic energy of an electron, $\alpha -$particle and a proton are given as $4K, 2K$ and $K$ respectively. The de-Broglie wavelength associated with electron $\left({\lambda }_e\right), \alpha -$particle $\left({\lambda }_{\alpha }\right)$ and the proton $\left({\lambda }_p\right)$ are as follows:

If de-Broglie wavelength is $\lambda$ when energy is $E$ (Kinetic energy). Find the wavelength at $\frac{E}{4}$(Kinetic energy).

Find ratio of de-Broglie wavelength of a proton and an $\alpha -$particle, when accelerated through a potential difference of $2 \mathrm{V}$ and $4 \mathrm{V}$ respectively.

For an electron and a proton$\left(m_p=1847 m_e\right)$ with same de-Broglie wavelength, the ratio of linear momentum is equal to:

Kinetic energy of electron, proton and a particle is given as $K, 2K$ and $4K$ respectively, then which of the following gives the correct order of de-Broglie wavelengths of electron, proton and a particle

Given below are two charged subatomic particles P and Q, that are accelerated through same potential difference $V$.

Here, Masses: $m_P=m_Q$ Charges: $\frac{1}{2}{q}_P=q_Q$

Which of the two sub atomic particles will have longer de Broglie wavelength?

de-Broglie wavelength associated with an electron in the $3^{\text{rd }}$ orbit of ${\mathrm{He}}^{+}$ ion is equal to

A subatomic particle of mass ${10}^{-18}\mu \mathrm{g}$ is in thermal equilibrium with its surrounding at a temperature of $400 \mathrm{K}$. Then the wave length of this particle will be

When the electron orbiting in hydrogen atom in its ground state moves to third excited state, the de-Broglie wavelength associated with it

A particle of mass $4\times {10}^{-27} \mathrm{kg}$ is moving with velocity $3\times {10}^5 \mathrm{m} {\mathrm{s}}^{-1}$. The de Broglie wavelength associated with the particle is(Use $h=6.6\times {10}^{-34} \mathrm{J} \mathrm{s}$)

If the mass and the kinetic energy of the particle is increased each by $1\%$, the percentage change in the de -Broglie wavelength is

In a photoelectric experiment light of wavelength $800 \mathrm{nm}$ produces photoelectrons with the smallest de Broglie wavelength of $1 \mathrm{nm}$. Then the work function of the metal used in the experiment is nearly

The momenta of a proton, a neutron and an electron are in the ratio $3:2:1$, then their respective de Broglie wavelengths are in the ratio

The ratio of de-Broglie wavelength associated with an electron in ground state and second excited state of a hydrogen atom is

Choose the correct statement.

Relation between mass and Planck constant