Question 132: Atoms & Nuclei

Question

The minimum frequency of photon required to break a particle of mass 15.348 amu into $4 \alpha$ particles is $\_\_\_\_$ kHz .
[mass of He nucleus = $4.002 \mathrm{amu}, 1 \mathrm{amu}=1.66 	imes 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.6 	imes 10^{-34} \mathrm{~J} . \mathrm{s}$ and $\mathrm{c}=3 	imes 10^8 \mathrm{~m} / \mathrm{s}$ ]

Accepted Answer

Einstein's equation tells us that mass and energy are interchangeable.
$$ \mathrm{E}=\mathrm{mc}^2 $$
When a nuclear reaction occurs, the total mass of the products is often slightly different from the initial mass of the reactants.
If the products are lighter, the missing mass is released as energy.
If the products are heavier, energy must be supplied to create that extra mass.
It is given that a particle breaks into $4 \alpha$ particles.
Initial mass of the single particle. $\mathrm{M}_{	ext {initial }}=15.348 \mathrm{amu}$
Final mass of $4 \alpha$ particles $=\mathrm{M}_{	ext {final }}=4 	imes$ mass of He nucleus
$$ \mathrm{M}_{\mathrm{final}}=4 	imes 4.002 \mathrm{amu}=16.008 \mathrm{amu} $$
The mass defect ( $\Delta \mathrm{m}$ ) during this nuclear reaction is,
$$ \Delta \mathrm{m}=\mathrm{M}_{	ext {final }}-\mathrm{M}_{	ext {initial }} $$
$$\Rightarrow $$ $\Delta \mathrm{m}=16.008 \mathrm{amu}-15.348 \mathrm{amu}$
$$\Rightarrow $$ $$ \Delta \mathrm{m}=0.66 \mathrm{amu} $$
It is given that $1 \mathrm{amu}=1.66 	imes 10^{-27} \mathrm{~kg}$.
$$ \Delta \mathrm{m}=0.66 	imes\left(1.66 	imes 10^{-27}ight) \mathrm{kg} $$
$$\Rightarrow $$ $$ \Delta \mathrm{m}=1.0956 	imes 10^{-27} \mathrm{~kg} $$
Now using Einstein's equation,
$\mathrm{E}=\Delta \mathrm{m} \cdot \mathrm{c}^2$
$$\Rightarrow $$ $E=\left(1.0956 	imes 10^{-27}ight) 	imes\left(3 	imes 10^8ight)^2$
$$\Rightarrow $$ $E=\left(1.0956 	imes 10^{-27}ight) 	imes\left(9 	imes 10^{16}ight)$
$$\Rightarrow $$ $$ E=9.8604 	imes 10^{-11} 	ext { Joules } $$
This energy is provided by a single photon. The energy of a photon is directly proportional to its frequency ( $v$ ), given by Planck's equation:
$$ \mathrm{E}=\mathrm{h} 
u $$
Where h is Planck's constant $\left(6.6 	imes 10^{-34} \mathrm{~J} \cdot \mathrm{~s}ight)$.
$$ v=\frac{E}{h} $$
$$\Rightarrow $$ $v=\frac{9.8604 	imes 10^{-11}}{6.6 	imes 10^{-34}}$
$$\Rightarrow $$$$ v=1.494 	imes 10^{23} \mathrm{~Hz} $$
$$\Rightarrow $$$$ v_{\mathrm{kHz}}=1.494 	imes 10^{23} 	imes 10^{-3} \mathrm{kHz} $$
$$\Rightarrow $$ $$ v_{\mathrm{kHz}}=14.94 	imes 10^{19} \mathrm{kHz} $$
Therefore, the minimum frequency of photon required is $14.94 	imes 10^{19} \mathrm{kHz}$.
Hence, the correct option is (D).

Answer

$14.94 	imes 10^{20}$

Answer

$9 	imes 10^{19}$

Answer

$9 	imes 10^{20}$

Answer

$14.94 	imes 10^{19}$

Atoms & Nuclei

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