Answer :
Certainly! Let's go through the calculation in detail.
We are given:
- Energy ([tex]\( E \)[/tex]) of the wave: [tex]\( 2.64 \times 10^{-21} \)[/tex] joules
- Planck's constant ([tex]\( h \)[/tex]): [tex]\( 6.6 \times 10^{-34} \)[/tex] joule-seconds
To find the frequency ([tex]\( f \)[/tex]) of the wave, we use Planck's equation:
[tex]\[ E = h \cdot f \][/tex]
We need to solve for [tex]\( f \)[/tex]:
[tex]\[ f = \frac{E}{h} \][/tex]
Substitute the given values into the equation:
[tex]\[ f = \frac{2.64 \times 10^{-21}}{6.6 \times 10^{-34}} \][/tex]
Now, perform the division:
[tex]\[ f = 4.00 \times 10^{12} \, \text{hertz} \][/tex]
So, the frequency of the wave is:
[tex]\[ f = 4.00 \times 10^{12} \, \text{hertz} \][/tex]
Therefore, the correct answer is:
A. [tex]\( 4.00 \times 10^{12} \)[/tex] hertz
We are given:
- Energy ([tex]\( E \)[/tex]) of the wave: [tex]\( 2.64 \times 10^{-21} \)[/tex] joules
- Planck's constant ([tex]\( h \)[/tex]): [tex]\( 6.6 \times 10^{-34} \)[/tex] joule-seconds
To find the frequency ([tex]\( f \)[/tex]) of the wave, we use Planck's equation:
[tex]\[ E = h \cdot f \][/tex]
We need to solve for [tex]\( f \)[/tex]:
[tex]\[ f = \frac{E}{h} \][/tex]
Substitute the given values into the equation:
[tex]\[ f = \frac{2.64 \times 10^{-21}}{6.6 \times 10^{-34}} \][/tex]
Now, perform the division:
[tex]\[ f = 4.00 \times 10^{12} \, \text{hertz} \][/tex]
So, the frequency of the wave is:
[tex]\[ f = 4.00 \times 10^{12} \, \text{hertz} \][/tex]
Therefore, the correct answer is:
A. [tex]\( 4.00 \times 10^{12} \)[/tex] hertz