A photon, an electron and a uranium nucleus all have the same wavelength. The one with the most energy
A photon, an electron and a uranium nucleus all have the same wavelength. The one with the most energy
- A
Is the photon
- B
Is the electron
- C
Is the uranium nucleus
- D
Depends upon the wavelength and the properties of the particle.
Similar Questions
Match the column
$(A)$ Hallwachs $\&$ Lenard
$(P)$ Transformers
$(B)$ Franck-Hertz
$(Q)$ Microwave
$(C)$ Klystron valve
$(R)$ Quantization of energy levels
$(D)$ Nicola Tesla
$(S)$ Photoelectric effect
Match the column
| $(A)$ Hallwachs $\&$ Lenard | $(P)$ Transformers |
| $(B)$ Franck-Hertz | $(Q)$ Microwave |
| $(C)$ Klystron valve | $(R)$ Quantization of energy levels |
| $(D)$ Nicola Tesla | $(S)$ Photoelectric effect |
The approximate wavelength of a photon of energy $2.48 \,eV$ is ............ $\mathring A$
A source $S_1$ is producing, $10^{15}$ photons per second of wavelength $5000 \;\mathring A.$ Another source $S_2$ is producing $1.02 \times 10^{15}$ photons per second of wavelength $5100\;\mathring A$. Then, $($ power of $S_2)/$ $($ power of $S_1)$ is equal to
A source $S_1$ is producing, $10^{15}$ photons per second of wavelength $5000 \;\mathring A.$ Another source $S_2$ is producing $1.02 \times 10^{15}$ photons per second of wavelength $5100\;\mathring A$. Then, $($ power of $S_2)/$ $($ power of $S_1)$ is equal to
- [AIPMT 2010]
There are ${n_1}$ photons of frequency ${\gamma _1}$ in a beam of light. In an equally energetic beam, there are ${n_2}$ photons of frequency ${\gamma _2}$. Then the correct relation is
There are ${n_1}$ photons of frequency ${\gamma _1}$ in a beam of light. In an equally energetic beam, there are ${n_2}$ photons of frequency ${\gamma _2}$. Then the correct relation is
$(i)$ In the explanation of photoelectric effect, we assume one photon of frequency v collides with an electron and transfers its energy. This leads to the equation for the maximum energy Emax of the emitted electron as $E_{max} = hf - \phi _0$ (where $\phi _0$ where do is the work function of the metal. If an electron absorbs $2$ photons (each of frequency $v$) what will be the maximum energy for the emitted electron ?
$(ii)$ Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential ?
$(i)$ In the explanation of photoelectric effect, we assume one photon of frequency v collides with an electron and transfers its energy. This leads to the equation for the maximum energy Emax of the emitted electron as $E_{max} = hf - \phi _0$ (where $\phi _0$ where do is the work function of the metal. If an electron absorbs $2$ photons (each of frequency $v$) what will be the maximum energy for the emitted electron ?
$(ii)$ Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential ?