The work function of caesium is 2.14 eV. The wavelength of incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V will be:

Photons absorbed in matter are converted to heat. A source emitting n photon/sec of frequency $\mathrm{\nu }$ is used to convert 1 kg of ice at $0°\mathrm{C}$ to water at $0°\mathrm{C}$. Then, the time T taken for the conversion:

(a) decreases with increasing n, with $\mathrm{\nu }$ fixed
(b) decreases with n fixed, $\mathrm{\nu }$ increasing
(c) remains constant with n and $\mathrm{\nu }$ changing such that n$\mathrm{\nu }$=constant
(d) increases when the product n$\mathrm{\nu }$ increases

The de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron. The speed of the electron is ${\mathrm{v}}_{\mathrm{e}}=\frac{\mathrm{c}}{100}$. Then,

Two particles A₁ and A₂ of masses m₁, m₂ (m₁>m₂) have the same de-Broglie wavelength. Then,

(a) their momenta are the same
(b) their energies are the same
(c) energy of A₁ is less than the energy of A₂
(d) energy of A₁ is more than the energy of A₂

Relativistic corrections become necessary when the expression for the kinetic energy $\frac{1}{2}{\mathrm{mv}}^2$, becomes comparable with ${\mathrm{mc}}^2$, where m is the mass of the particle. At what de-Broglie wavelength, will relativistic corrections become important for an electron?

(a) $\mathrm{\lambda }=10 \mathrm{nm}$
(b) $\mathrm{\lambda }={10}^{-1} \mathrm{nm}$
(c) $\mathrm{\lambda }={10}^{-4} \mathrm{nm}$
(d) $\mathrm{\lambda }={10}^{-6} \mathrm{nm}$

An electron (mass m) with an initial velocity $\overrightarrow{\mathrm{v}}={\mathrm{v}}_0\hat{\mathrm{i}}$ is in an electric field $\overrightarrow{\mathrm{E}}={\mathrm{E}}_0\hat{\mathrm{j}}$. If ${\mathrm{\lambda }}_0=\frac{\mathrm{h}}{{\mathrm{mv}}_0}$, its de-Broglie wavelength at time t is given by:

An electron (mass m) with an initial velocity $\overrightarrow{\mathrm{v}}={\mathrm{v}}_0\hat{\mathrm{i}} ({\mathrm{v}}_0>0)$ is in an electric field $\overrightarrow{\mathrm{E}}=-{\mathrm{E}}_0\hat{\mathrm{i}} ({\mathrm{E}}_0=\mathrm{constant}>0)$. Its de-Broglie wavelength at time t is given by:

An electron is moving with an initial velocity $\overrightarrow{\mathrm{v}}={\mathrm{v}}_0\hat{\mathrm{i}}$ and is in a magnetic field $\overrightarrow{\mathrm{B}}={\mathrm{B}}_0\hat{\mathrm{j}}$. Then, its de-Broglie wavelength:

A proton, a neutron, an electron and an $\mathrm{\alpha }$-particle have the same energy. Then, their de-Broglie wavelengths compare as:

Monochromatic light of frequency $6.0\times {10}^{14} \mathrm{Hz}$ is produced by a laser. The power emitted is $2.0\times {10}^{-3} \mathrm{W}.$ How many photons per second, on average, are emitted by the source?

$$\begin{gathered} 1. 6\times {10}^{15} \\ 2. 5\times {10}^{14} \\ 3. 6\times {10}^{14} \\ 4. 5\times {10}^{15} \end{gathered}$$

The wavelength of light in the visible region is about 390 nm for violet colour. The energy of photons in (eV) is:

$\left(\mathrm{Take} \mathrm{h}=6.63\times {10}^{-34} \mathrm{Js}, \mathrm{and} 1 \mathrm{eV}=1.6\times {10}^{-19} \mathrm{J}\right)$

What is the de-Broglie wavelength associated with an electron moving at a speed of $5.4\times {10}^6 \mathrm{m}/\mathrm{s}?$

The de Broglie wavelength associated with a ball of mass 150 g travelling at a speed of 30.0 m/s is:

$$\begin{gathered} 1. 1.47\times {10}^{-32} \mathrm{m} \\ 2. 2.01\times {10}^{-34} \mathrm{m} \\ 3. 2.01\times {10}^{-32} \mathrm{m} \\ 4. 1.47\times {10}^{-34} \mathrm{m} \end{gathered}$$

An electron, an α-particle, and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?

A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is $1.813\times {10}^{-4}$. The particle’s mass is:

$$\begin{gathered} 1. 9.371\times {10}^{-31} \mathrm{kg} \\ 2. 9.001\times {10}^{-31} \mathrm{kg} \\ 3. 1.675\times {10}^{-27} \mathrm{kg} \\ 4. 2.705\times {10}^{-27} \mathrm{kg} \end{gathered}$$

What is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts?

Graph shows variation of photoelectric current with intensity of light. It can be concluded that photoelectrons emitted per second are-

The graph shows the variation of photocurrent with collector plate potential for different intensities of incident radiation. Then the maximum intensity is:

The wavelength of light in the visible region is about 550 nm (average wavelength) for yellow-green colour. Three materials with work functions are given as Al (4.28 eV), Cu (4.65 eV) and Na (2.75 eV). From which of these photosensitive materials, can you build a photoelectric device that operates with visible light?

The number of photons per second on an average emitted by a source of monochromatic light of wavelength 600 nm, when it delivers the power of $3.3\times {10}^{-3}$ watt will be: $\left(\mathrm{h}=6.6\times {10}^{-34} \mathrm{Js}\right)$

An electromagnetic wave of wavelength $\text{'}\mathrm{\lambda }\text{'}$ is incident on a photosensitive surface of negligible work function. If 'm' is mass of photoelectron emitted from the surface has de-Broglie wavelength ${\mathrm{\lambda }}_{\mathrm{d}}$, then: