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Chapter 7: Problem 57

It takes \(7.21 \times 10^{-19} \mathrm{J}\) of energy to remove an electron from an iron atom. What is the maximum wavelength of light that can do this?

Short Answer

Expert verified
The maximum wavelength of light that can remove an electron from an iron atom is \(2.76 \times 10^{-7}\) meters.

Step by step solution

01

Understand the relationship between energy and wavelength

The energy of a photon can be calculated using the equation: \( E = h\cdot f \) where E is the energy, h is Planck's constant (approximately \(6.63 \times 10^{-34} \mathrm{Js}\)), and f is the frequency of the photon. Since the frequency is related to the speed of light \( c \) by the equation and the wavelength \(\lambda\): \(f = c / \lambda \), We can rewrite the energy equation in terms of wavelength as follows: \( E = h\cdot c / \lambda \) We are given the energy required to remove an electron from an iron atom and asked to find the maximum wavelength of light that can do this, so we will solve the equation for the wavelength \(\lambda\):
02

Solve the energy equation for wavelength

Now we need to solve the equation for \(\lambda\). To isolate \(\lambda\), we will first divide both sides by \(h \cdot c\): \( \frac{E}{h\cdot c} = \frac{1}{\lambda} \) Now, we can take the reciprocal of both sides to find the value of \(\lambda\): \( \lambda = \frac{1}{\frac{E}{h\cdot c}} \)
03

Plug in the given energy value and constants

Using the given energy value, and the constants for Planck's constant and the speed of light, we can now calculate the wavelength: \( \lambda = \frac{1}{\frac{7.21 \times 10^{-19} \mathrm{J}}{(6.63 \times 10^{-34} \mathrm{Js}) \cdot (3.00 \times 10^{8} \, \mathrm{m/s})}} \)
04

Calculate the maximum wavelength

Now, we perform the calculations and find the maximum wavelength of light that can remove an electron from an iron atom: \( \lambda = \frac{1}{\frac{7.21 \times 10^{-19} \mathrm{J}}{(6.63 \times 10^{-34} \mathrm{Js}) \cdot (3.00 \times 10^{8} \, \mathrm{m/s})}} \) \( \lambda = \frac{1}{\frac{7.21 \times 10^{-19}}{1.989 \times 10^{-25}}} \) \( \lambda = \frac{1}{3.62 \times 10^6 \, \mathrm{m^{-1}}} \) \( \lambda = 2.76 \times 10^{-7} \, \mathrm{m} \) So, the maximum wavelength of light that can remove an electron from an iron atom is \(2.76 \times 10^{-7}\) meters.

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Most popular questions from this chapter

Does a photon of visible light \((\lambda \approx 400 \text { to } 700 \mathrm{nm})\) have sufficient energy to excite an electron in a hydrogen atom from the \(n=1\) to the \(n=5\) energy state? from the \(n=2\) to the \(n=6\) energy state?

In the ground state of mercury, Hg, a. how many electrons occupy atomic orbitals with \(n=3 ?\) b. how many electrons occupy \(d\) atomic orbitals? c. how many electrons occupy \(p_{z}\) atomic orbitals? d. how many electrons have spin "up" " \(\left(m_{s}=+\frac{1}{2}\right) ?\)

Give the name and formula of each of the binary compounds formed from the following elements. a. Li and N b. Na and Br c. K and S

Give a possible set of values of the four quantum numbers for all the electrons in a boron atom and a nitrogen atom if each is in the ground state.

The electron affinities of the elements from aluminum to chlorine are \(-44,-120,-74,-200.4,\) and \(-384.7 \mathrm{kJ} / \mathrm{mol}\) respectively. Rationalize the trend in these values.
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