Question

Units of Planck’s constant in CGS system are:
A) Erg per second
B) Second per erg
C) Erg second
D) Erg per second per second

Answer 1

Hint: Use the relation of Planck’s constant , energy and frequency to determine the unit. This relation is also called einstein energy formula for quantisation of energy.

Formula Used:
The formula that we will use for determining the unit of Planck’s constant in CGS unit can be written as $E = hf$ where $E$ represents energy, $h$ represents Planck’s constant and $f$ represents frequency.

Complete step by step answer:
We can find out the unit of Planck’s constant using two approaches. We will see them one by one.
The first method will be a mathematical approach in which we will be using the formula $E = hf$ for calculating the unit.
So here we will first write the equation $E = hf$
Then our next step is to write the equation in a way such that we have $h$ on LHS of the equation and $E$ and $f$ on the RHS of the equation. So, we may rewrite the previous equation as,
$h = \dfrac{{\text{E}}}{{\text{f}}}$
Now we already know that the CGS unit of energy is erg and that of frequency(which is inverse of time) is ${s^{ - 1}}$(i.e. second inverse)
So, we can write that, $h = \frac{{{\text{erg}}}}{{{{\text{s}}^{ - 1}}}}$
Or, $h = erg \times s$
Hence the unit of Planck’s constant is Erg second (option C).
The second approach can be used if we already know the SI unit of Planck’s constant. So, Planck’s constant in SI unit is given as $h = 6.62607004 \times {10^{ - 34}}Js$ where $J$ represents joule and $s$ represents second.
Now individually if we convert the SI units to CGS units we have $J \to erg$ and $s \to s$

So, we get the CGS unit as Erg second (option C).

Note: It is always better to use SI units for solving sums. Unless it is stated in the question to use another unit system one must always use SI units. The value of Planck’s constant in different unit systems are:
$h = 6.62607004 \times {10^{ - 34}}Js$ (SI unit)
$h = 6.62607004 \times {10^{ - 29}}erg/second$ (CGS unit)
Usually for solving sums if the value of the constant is not given it is best to take the value as $h = 6.626 \times {10^{ - 34}}Js$.