142400 A particle of mass $2 \times 10^{-27} \mathrm{~kg}$ has de-Broglie wavelength of $3.3 \times 10^{-10} \mathrm{~m}$. The kinetic energy of this particle is
(Plank's Constant $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )

142401 A nucleus of mass $M$ at rest splits into two parts having masses $\frac{M^{\prime}}{3}$ and $\frac{2 M^{\prime}}{3} \quad M^{\prime} \lt $ M).The ratio of de Broglie wavelength of two parts will be:

142402 The equation $\lambda=\frac{1.227}{\mathrm{x}} \mathrm{nm}$ can be used to
find the de-Broglie wavelength of an electron. In this equation $x$ stands for: where, $\mathbf{m}=$ mass of electron
$\mathrm{P}=$ momentum of electron
$K=$ Kinetic energy of electron
$V=$ Accelerating potential in volts for electron

142403 The de Broglie wavelengths for an electron and a photon are $\lambda_{e}$ and $\lambda_{p}$ respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two?

142404 An $\alpha$ particle and a carbon 12 atom has same kinetic energy $K$. The ratio of their de-Broglie wavelengths $\left(\lambda \alpha: \lambda_{\mathrm{C} 12}\right)$ is :

142400 A particle of mass $2 \times 10^{-27} \mathrm{~kg}$ has de-Broglie wavelength of $3.3 \times 10^{-10} \mathrm{~m}$. The kinetic energy of this particle is
(Plank's Constant $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )

142401 A nucleus of mass $M$ at rest splits into two parts having masses $\frac{M^{\prime}}{3}$ and $\frac{2 M^{\prime}}{3} \quad M^{\prime} \lt $ M).The ratio of de Broglie wavelength of two parts will be:

142402 The equation $\lambda=\frac{1.227}{\mathrm{x}} \mathrm{nm}$ can be used to
find the de-Broglie wavelength of an electron. In this equation $x$ stands for: where, $\mathbf{m}=$ mass of electron
$\mathrm{P}=$ momentum of electron
$K=$ Kinetic energy of electron
$V=$ Accelerating potential in volts for electron

142403 The de Broglie wavelengths for an electron and a photon are $\lambda_{e}$ and $\lambda_{p}$ respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two?

142404 An $\alpha$ particle and a carbon 12 atom has same kinetic energy $K$. The ratio of their de-Broglie wavelengths $\left(\lambda \alpha: \lambda_{\mathrm{C} 12}\right)$ is :

142400 A particle of mass $2 \times 10^{-27} \mathrm{~kg}$ has de-Broglie wavelength of $3.3 \times 10^{-10} \mathrm{~m}$. The kinetic energy of this particle is
(Plank's Constant $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )

142401 A nucleus of mass $M$ at rest splits into two parts having masses $\frac{M^{\prime}}{3}$ and $\frac{2 M^{\prime}}{3} \quad M^{\prime} \lt $ M).The ratio of de Broglie wavelength of two parts will be:

142402 The equation $\lambda=\frac{1.227}{\mathrm{x}} \mathrm{nm}$ can be used to
find the de-Broglie wavelength of an electron. In this equation $x$ stands for: where, $\mathbf{m}=$ mass of electron
$\mathrm{P}=$ momentum of electron
$K=$ Kinetic energy of electron
$V=$ Accelerating potential in volts for electron

142403 The de Broglie wavelengths for an electron and a photon are $\lambda_{e}$ and $\lambda_{p}$ respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two?

142404 An $\alpha$ particle and a carbon 12 atom has same kinetic energy $K$. The ratio of their de-Broglie wavelengths $\left(\lambda \alpha: \lambda_{\mathrm{C} 12}\right)$ is :

142400 A particle of mass $2 \times 10^{-27} \mathrm{~kg}$ has de-Broglie wavelength of $3.3 \times 10^{-10} \mathrm{~m}$. The kinetic energy of this particle is
(Plank's Constant $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )

142401 A nucleus of mass $M$ at rest splits into two parts having masses $\frac{M^{\prime}}{3}$ and $\frac{2 M^{\prime}}{3} \quad M^{\prime} \lt $ M).The ratio of de Broglie wavelength of two parts will be:

142402 The equation $\lambda=\frac{1.227}{\mathrm{x}} \mathrm{nm}$ can be used to
find the de-Broglie wavelength of an electron. In this equation $x$ stands for: where, $\mathbf{m}=$ mass of electron
$\mathrm{P}=$ momentum of electron
$K=$ Kinetic energy of electron
$V=$ Accelerating potential in volts for electron

142403 The de Broglie wavelengths for an electron and a photon are $\lambda_{e}$ and $\lambda_{p}$ respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two?

142404 An $\alpha$ particle and a carbon 12 atom has same kinetic energy $K$. The ratio of their de-Broglie wavelengths $\left(\lambda \alpha: \lambda_{\mathrm{C} 12}\right)$ is :

142400 A particle of mass $2 \times 10^{-27} \mathrm{~kg}$ has de-Broglie wavelength of $3.3 \times 10^{-10} \mathrm{~m}$. The kinetic energy of this particle is
(Plank's Constant $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )

142401 A nucleus of mass $M$ at rest splits into two parts having masses $\frac{M^{\prime}}{3}$ and $\frac{2 M^{\prime}}{3} \quad M^{\prime} \lt $ M).The ratio of de Broglie wavelength of two parts will be:

142402 The equation $\lambda=\frac{1.227}{\mathrm{x}} \mathrm{nm}$ can be used to
find the de-Broglie wavelength of an electron. In this equation $x$ stands for: where, $\mathbf{m}=$ mass of electron
$\mathrm{P}=$ momentum of electron
$K=$ Kinetic energy of electron
$V=$ Accelerating potential in volts for electron

142403 The de Broglie wavelengths for an electron and a photon are $\lambda_{e}$ and $\lambda_{p}$ respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two?

142404 An $\alpha$ particle and a carbon 12 atom has same kinetic energy $K$. The ratio of their de-Broglie wavelengths $\left(\lambda \alpha: \lambda_{\mathrm{C} 12}\right)$ is :