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Photon is quantum of radiation with energy $E=h \mathrm{v}$, where $v$ is frequency and $h$ is Planck's constant. The dimensions of $h$ are the same as that of
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Verified Answer
The correct answers are:
angular impulse
,
angular momentum
angular impulse
,
angular momentum
As we know that
Energy of radiation, $E=h \mathrm{v}$
$$
[h]=\frac{[E]}{[\mathrm{v}]}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{T}^{-1}\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]
$$
$\begin{aligned} \text { Linear impulse } &=F t=\frac{d P}{d t} \times d t \\ &=d P=m v=\left[\mathrm{MLT}^{-1}\right] \end{aligned}$
So, dimension of linear impulse
$=$ Dimension of momentum $=\left[\mathrm{MLT}^{-1}\right]$
And, angular impulse
$$
=\tau d t=\frac{d L}{d t} \times d t=d L
$$
$=$ Change in angular momentum.
$d L=m v r=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$
So, dimension of angular impulse
$=$ Dimension of angular momentum
$=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$
Hence, this is similar to the dimension of Planck's constant $h$.
Energy of radiation, $E=h \mathrm{v}$
$$
[h]=\frac{[E]}{[\mathrm{v}]}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{T}^{-1}\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]
$$
$\begin{aligned} \text { Linear impulse } &=F t=\frac{d P}{d t} \times d t \\ &=d P=m v=\left[\mathrm{MLT}^{-1}\right] \end{aligned}$
So, dimension of linear impulse
$=$ Dimension of momentum $=\left[\mathrm{MLT}^{-1}\right]$
And, angular impulse
$$
=\tau d t=\frac{d L}{d t} \times d t=d L
$$
$=$ Change in angular momentum.
$d L=m v r=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$
So, dimension of angular impulse
$=$ Dimension of angular momentum
$=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$
Hence, this is similar to the dimension of Planck's constant $h$.
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