142544 A proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

142545 The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^{8} \mathrm{~m} / \mathrm{s}$ is equal to that of a photon. The ratio of kinetic energy of the electron to that of the photon $\left(c=3 \times 10^{8}\right.$ $\mathbf{m} / \mathbf{s})$

142546 The energy of a photon is equal to the kinetic energy of a proton. If $\lambda_{1}$ is the de-Broglie wavelength of a proton, $\lambda_{2}$ the wavelength associated with the photon and if the energy of the photon is $E$, then $\left(\lambda_{1} / \lambda_{2}\right)$ is proportional to

142548 A photo sensitive metallic surface emits electrons when $X$-rays of wavelength $\lambda$ fall on it. The de-Broglie wavelength of the emitted electrons is (Neglect the work function of the surface, $m$ is mass of the electron, $h$ is Planck's $c$ is the velocity of light

142544 A proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

142545 The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^{8} \mathrm{~m} / \mathrm{s}$ is equal to that of a photon. The ratio of kinetic energy of the electron to that of the photon $\left(c=3 \times 10^{8}\right.$ $\mathbf{m} / \mathbf{s})$

142546 The energy of a photon is equal to the kinetic energy of a proton. If $\lambda_{1}$ is the de-Broglie wavelength of a proton, $\lambda_{2}$ the wavelength associated with the photon and if the energy of the photon is $E$, then $\left(\lambda_{1} / \lambda_{2}\right)$ is proportional to

142548 A photo sensitive metallic surface emits electrons when $X$-rays of wavelength $\lambda$ fall on it. The de-Broglie wavelength of the emitted electrons is (Neglect the work function of the surface, $m$ is mass of the electron, $h$ is Planck's $c$ is the velocity of light

142544 A proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

142545 The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^{8} \mathrm{~m} / \mathrm{s}$ is equal to that of a photon. The ratio of kinetic energy of the electron to that of the photon $\left(c=3 \times 10^{8}\right.$ $\mathbf{m} / \mathbf{s})$

142546 The energy of a photon is equal to the kinetic energy of a proton. If $\lambda_{1}$ is the de-Broglie wavelength of a proton, $\lambda_{2}$ the wavelength associated with the photon and if the energy of the photon is $E$, then $\left(\lambda_{1} / \lambda_{2}\right)$ is proportional to

142548 A photo sensitive metallic surface emits electrons when $X$-rays of wavelength $\lambda$ fall on it. The de-Broglie wavelength of the emitted electrons is (Neglect the work function of the surface, $m$ is mass of the electron, $h$ is Planck's $c$ is the velocity of light

142544 A proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

142545 The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^{8} \mathrm{~m} / \mathrm{s}$ is equal to that of a photon. The ratio of kinetic energy of the electron to that of the photon $\left(c=3 \times 10^{8}\right.$ $\mathbf{m} / \mathbf{s})$

142546 The energy of a photon is equal to the kinetic energy of a proton. If $\lambda_{1}$ is the de-Broglie wavelength of a proton, $\lambda_{2}$ the wavelength associated with the photon and if the energy of the photon is $E$, then $\left(\lambda_{1} / \lambda_{2}\right)$ is proportional to

142548 A photo sensitive metallic surface emits electrons when $X$-rays of wavelength $\lambda$ fall on it. The de-Broglie wavelength of the emitted electrons is (Neglect the work function of the surface, $m$ is mass of the electron, $h$ is Planck's $c$ is the velocity of light