Question

What is the threshold velocity vthreshold(ethanol) for creating Cherenkov light from a charged particle as it travels through ethanol (which has an index of refraction of n

Answer 1

Explanation:

The velocity of light in a medium of refractive index $n[tex] is given by, [tex]v=\frac{c}{n}$

$v \text { is the velocity of light in the medium }$

$c \text { is speed of light in vacuum }$

The exact value of speed of light in vacuum is $299792458 \mathrm{m} / \mathrm{s}$.

For Cherenkov radiation to be emitted, the velocity of the charged particle traversing the medium must be greater than this velocity. Thus, the threshold velocity of for creating Cherenkov radiation is,

$v_{\text {Cherenkov }} \geq \frac{c}{n}$

$v_{\text {threshod }}=\frac{c}{n}$

For water $n=1.33,[tex] thus the threshold velocity for producing Cherenkov radiation in water is, [tex]v_{\text {threatold }(\text { water })} &=\frac{299792458 \mathrm{m} / \mathrm{s}}{1.33}$

$=225407863 \mathrm{m} / \mathrm{s}$

$=2.254 \times 10^{8} \mathrm{m} / \mathrm{s}$

For ethanol $n=1.36[tex], thus the threshold velocity for producing Cherenkov radiation in water is, [tex]v_{\text {threstold }( \text { ettanol) } } &=\frac{299792458 \mathrm{m} / \mathrm{s}}{1.36}$

$=220435630 \mathrm{m} / \mathrm{s}$

$=2.204 \times 10^{8} \mathrm{m} / \mathrm{s}$

Answer 2

Answer:

The answer is "2.2 × $\bold{10^8}$".

Explanation:

In the given question the value of n is missing which can be defined as follows:

n= 1.36

The velocity value of the threshold(ethanol) for a generation the Cerenkov light from the charged particle by travel through ethanol as:

know we will have to use an equation as follows:

Formula:      

(ethanol) or the vthreshold = $\frac{c}{n}$

                                         $= \frac{3\times 10^8} {1.36} \\\\= 2.2 \times 10^8$

The water in vthreshold:

$= 2.2 \times 10^8 \ \ \frac{m}{ s}$

Express the value in c, that is multiple, so, the value of vthreshold(water) is:

=(0.735) c