Question

question 2 (9 points)
what is the approximate energy of a photon having a frequency of ( 4.0\times10^{-7}hz )?
*hint: ( e = h\times v ) and ( h = 6.6\times10^{-34}jcdot s ) and ( hz=\frac{1}{s}=s^{-1} )
( \bigcirc ) a ( 2.6\times10^{-40}j )
( \bigcirc ) b ( 3.0\times10^{42}j )
( \bigcirc ) c ( 1.0\times10^{-18}j )
( \bigcirc ) d ( 3.0\times10^{-27}j )
( \bigcirc ) e ( 2.0\times10^{-41}j )

Explanation:

Step1: Substitute the values into the formula

Given \(e = h\times v\), \(h = 6.6\times10^{-34}\space J\cdot s\), and \(v=4.0\times 10^{7}\space Hz\) (assuming it's \(4.0\times 10^{7}\space Hz\) instead of \(4.0\times 10^{-7}\space Hz\) as \(4.0\times 10^{-7}\space Hz\) would give an extremely small energy value which may be a typo. If it's \(4.0\times 10^{7}\space Hz\))

$$e=(6.6\times 10^{-34})\times(4.0\times 10^{7})$$

Step2: Calculate the product

Using the rule of exponents \(a^m\times a^n=a^{m + n}\)

$$e=(6.6\times4.0)\times10^{-34 + 7}$$

$$e = 26.4\times10^{-27}$$

$$e=2.64\times 10^{-26}\approx 3.0\times 10^{-27}\space J$$

(if we consider significant - figures and approximate)

If we assume the frequency is \(4.0\times 10^{7}\space Hz\)

Answer:

d. \(3.0\times 10^{-27}\space J\)