$A$ scientist claims to have a perfect technique in which he can spontaneously convert an electron completely into energy in the laboratory without any other material required. What is the conclusion about this claim from our current understanding of physics?
- AThis is possible because Einstein's equation says that mass and energy are equivalent... it is just very difficult to achieve with electrons
- BThis is possible and it is done all the time in the high-energy physics labs.
- CThe scientist is almost correct... except that in converting the electron to energy, an electron's anti-particle is produced in the process as well
- DThis is not possible because charge conservation would be violated.
Explore More
Similar Questions
The minimum intensity of light that can be perceived by the human eye is $10^{-10} \ W/m^2$. If the area of the pupil is $10^{-6} \ m^2$ and the wavelength of light is $560 \ nm$,approximately how many photons per second enter the eye? (Use $h = 6.6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$).
Difficult
View SolutionIf the momentum of a photon is $p$,then its frequency is ........
Easy
View SolutionEstimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never 'count photons',even in barely detectable light.
$(a)$ The number of photons emitted per second by a Medium wave transmitter of $10\; kW$ power,emitting radiowaves of wavelength $500\; m$.
$(b)$ The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive $(10^{-10}\; W m^{-2})$. Take the area of the pupil to be about $0.4\; cm^2$,and the average frequency of white light to be about $6 \times 10^{14}\; Hz$.
$(a)$ The number of photons emitted per second by a Medium wave transmitter of $10\; kW$ power,emitting radiowaves of wavelength $500\; m$.
$(b)$ The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive $(10^{-10}\; W m^{-2})$. Take the area of the pupil to be about $0.4\; cm^2$,and the average frequency of white light to be about $6 \times 10^{14}\; Hz$.
Medium
View Solution$10^{20}$ photons of wavelength $660 \ nm$ are emitted per second from a lamp. The wattage of the lamp is (Planck's constant $h = 6.6 \times 10^{-34} \ J \cdot s$). (in $W$)
MediumWBJEE 2025
View SolutionWhen light of wavelength $\lambda$ is incident on a photosensitive surface,photons of power $P$ are emitted. The number of photons $n$ emitted in time $t$ is: [$h$ = Planck's constant,$c$ = velocity of light in vacuum]
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo