Question

The ratio of SI unit to the CGS unit of Planck’s constant is:
A) ${10^7}:1$
B) ${10^4}:1$
C) ${10^6}:1$
D) $1:1$

Answer 1

Hint: To solve this question, we need to know the basic theory related to the physical quantities with its dimension and proper units as some of them describe below. As we know Planck’s constant is a quantity with the dimensions of “${{energy \times time}}$” and comes in units like “\[{{Joules \times Second}}\]”. And by using these first we will get the Planck's constant (h) in SI unit and then in CGS unit and then after taking its ratio as described below.

Formula used:
${{E = hf}}$ and
${{h = }}\dfrac{{{{energy}}}}{{\dfrac{{{c}}}{{{\lambda }}}}}$
Where, E is energy, c is speed of light, h is Planck's constant and ${{\lambda }}$ is wavelength.

Complete step by step solution:
Planck's constant is represented by h in physics. And it relates the energy of one quantum (photon) electromagnetic radiation to the frequency of that radiation.
In the centimeter-gram-second (CGS) (which is also called small-unit metric system), it is equal to approximately ${{6}}{{.6 \times 1}}{{{0}}^{{{ - 27}}}}{{Erg second}}$.
The energy E contained in a light photon, which generally represents the smallest possible 'packet' of energy in an electromagnetic wave channel and it is directly proportional to the frequency (f) according to the below equation:
${{E = hf}}$
Where, E represents contained in a light photon,
h represents Planck's constant and
f represents frequency.
If f is given in hertz (the unit measure of frequency) and E is given in joules, then in this case we know it is equal to approximately ${{6}}{{.6 \times 1}}{{{0}}^{{{ - 27}}}}{{Erg second}}$.
And finally, we have,
$\Rightarrow$ ${{E = 6}}{{.6 \times 1}}{{{0}}^{{{ - 27}}}} \times {{f}}$
As we know,
$\Rightarrow$ ${{E = hf}}$
$\Rightarrow$ ${{h = }}\dfrac{{{{energy}}}}{{\dfrac{{{c}}}{{{\lambda }}}}}$
Where c is the speed of light, h is Planck's constant and ${{\lambda }}$ is wavelength.
Dimensional formula of energy and speed of light in above equation, we get
Energy = $\left[ {{{M}}{{{L}}^{{2}}}{{{T}}^{{{ - 2}}}}} \right]$and speed of light = $\left[ {{{L}}{{{T}}^{{{ - 1}}}}} \right]$
$\Rightarrow$ ${{h = }}\dfrac{{\left[ {{{M}}{{{L}}^{{2}}}{{{T}}^{{{ - 2}}}}} \right]}}{{\dfrac{{\left[ {{{L}}{{{T}}^{{{ - 1}}}}} \right]}}{{\left[ {{L}} \right]}}}}$
$\Rightarrow$ ${{h = M}}{{{L}}^{{2}}}{{{T}}^{{{ - 1}}}}$
Thus, h in SI unit is ${{kg}}\dfrac{{{{{m}}^{{2}}}}}{{{s}}}$.
And h in c.g.s unit = $\left( {{{1}}{{{0}}^{{3}}}{{gm}}} \right){{ \times }}{\left( {{{1}}{{{0}}^{{2}}}} \right)^{{2}}}\dfrac{{{{c}}{{{m}}^{{2}}}}}{{{s}}}$ = ${{1}}{{{0}}^{{7}}}{{gm}}\dfrac{{{{c}}{{{m}}^{{2}}}}}{{{s}}}$
Now, we have to calculate the ratio of SI units to the CGS unit.
Ratio= $\dfrac{{\left( {{{1}}{{{0}}^{{3}}}} \right){{ \times }}{{\left( {{{1}}{{{0}}^{{2}}}} \right)}^{{2}}}}}{1}$= ${{1}}{{{0}}^{{7}}}:1$
Thus, the ratio of SI unit to the CGS unit is ${{1}}{{{0}}^{{7}}}:1$.

Therefore, option (A) is the correct answer.

Note: Always remember that SI unit is same everywhere as in cgs unit there are different units of measurement but SI unit is adopted internationally and the units are fixed and do not change at any place.