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A clean iron surface is illuminated by ultraviolet light. No photoelectrons are ejected until the wavelength of the incident UV light falls below 288 nm. (a) What is the work function (in electron-volts) of the metal? (b) What is the maximum kinetic energy for electrons ejected by incident light of wavelength \(140 \mathrm{nm} ?\)

Short Answer

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#tag_title#Step 2: Calculate the work function#tag_content# Now that we have the threshold frequency, we can calculate the work function using the Planck-Einstein relation. The work function (桅) is given by 桅 = h 脳 谓_threshold, where h is the Planck's constant (6.63 脳 10鈦宦斥伌 Js) and 谓_threshold is the threshold frequency. Calculate the work function: $$\Phi = h \cdot \nu_{threshold}$$ #tag_title#Step 3: Find the maximum kinetic energy#tag_content# We have to find the maximum kinetic energy of the electrons when the incident light has a wavelength of 140 nm. Given wavelength: 位 = 140 nm = 140 脳 10鈦烩伖 m First, we need to find the frequency (谓) of the incident light: $$\nu = \frac{c}{\lambda}$$ Next, we will use the photoelectric equation to find the maximum kinetic energy (K_max): $$K_{max} = h \cdot \nu - \Phi$$ Now, we can plug in all the known values and calculate the maximum kinetic energy (K_max).

Step by step solution

01

Find the threshold frequency

To find the threshold frequency, we will first convert the threshold wavelength to frequency using the speed of light formula, where \(c = \lambda_{threshold} \cdot \nu_{threshold}\), with \(c\) representing the speed of light, \(\lambda_{threshold}\) being the threshold wavelength and \(\nu_{threshold}\) being the threshold frequency. Given wavelength: \(\lambda_{threshold} = 288 \mathrm{nm} = 288 \times 10^{-9} \mathrm{m}\) The speed of light, c = \(3 \times 10^8 \mathrm{m/s}\) Calculate the threshold frequency: $$\nu_{threshold} = \frac{c}{\lambda_{threshold}}$$

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