The energy of a gamma (gamma) ray photon is E | vedclass

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Question

(C) The energy of a photon is given by $E = h
u = \frac{hc}{\lambda}$. Since the frequency $
u$ of electromagnetic waves increases in the order: Vis

Vedclass · Accepted Answer

$E_{\gamma} > E_{X} > E_{v}$

Vedclass · Answer

(C) The energy of a photon is given by $E = h
u = \frac{hc}{\lambda}$. Since the frequency $
u$ of electromagnetic waves increases in the order: Visible light $1$. For visible light, the energy range is approximately $1.6 \ eV$ to $3.2 \ eV$ $(E_{v} \approx 2.48 \ eV)$.$2$. For $X$-rays, the energy range is approximately $100 \ eV$ to $100 \ keV$.$3$. For gamma rays, the energy is typically greater than $100 \ keV$.Comparing these ranges, we find that $E_{\gamma} > E_{X} > E_{v}$.

Part of the electromagnetic spectrum	Applications
$P$. Microwave	$I$. For purifying the water
$Q$. $UV$ rays	$II$. For warming the food
$R$. Gamma rays	$III$. For $AM$ and $FM$ communication systems
$S$. Radio wave	$IV$. For treating the cancer cells

The energy of a gamma $(\gamma)$ ray photon is $E_{\gamma}$, and that of an $X$-ray photon is $E_{X}$. If the visible light photon has an energy of $E_{v}$, then we can say that:

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