
Concept explainers
(a)
The longest wavelength (in nanometer) that can dislodge an electron from a negative hydrogen ion. Given that the amount of energy (
(a)

Answer to Problem 33Q
Solution:
Explanation of Solution
Given data:
The amount of energy (
Formula used:
Write the expression for the energy of a photon:
Here,
Explanation:
As Planck’s constant,
Also consider the speed of light,
Refer to the expression for the energy of a photon.
Substitute
Conclusion:
Hence, the longest wavelength that will emit an electron from the negative hydrogen ion will be
(b)
The part of the
(b)

Answer to Problem 33Q
Solution:
This wavelength will lie in the infrared region of the electromagnetic spectrum.
Explanation of Solution
Introduction:
The electromagnetic spectrum is broadly classified into three classes, which are ultraviolet region, visible, and infrared regions. The wavelength increases while the energy decreases from the ultraviolet to infrared region. The visible region lies between 400 to 700 nm.
Explanation:
The wavelength calculated in part (a) came out to be
Conclusion:
Therefore, the wavelength that will emit an electron from the negative hydrogen ion will lie in the infrared region of the electromagnetic spectrum.
(c)
Whether a photon of visible light will be able to dislodge the same electron from a negative hydrogen ion. Given that the amount of energy (
(c)

Answer to Problem 33Q
Solution:
Yes, a photon of the visible light will be able to dislodge an extra electron.
Explanation of Solution
Introduction:
The visible light is called so because it can be seen through naked eyes. Wavelength other than from the visible region will only be seen through special filters. The visible part lies between the infrared and ultraviolet regions in the electromagnetic spectrum.
Explanation:
The energy level of the visible light is higher than that of the infrared radiations. As the minimum energy required to expel an electron lies in the infrared region, a photon of visible light will definitely be able to do so. The photon of visible light will be able to excite the electron much quicker than that of the infrared region.
Conclusion:
Therefore, a photon of the visible region will be able to excite an electron from the negative hydrogen ion much quickly than by the one lying in the infrared region.
(d)
The reason for the opacity of the photosphere that contains negative hydrogen ions for the visible light, but not for the infrared light. Given that the amount of energy (
(d)

Answer to Problem 33Q
Solution:
The photosphere is opaque (invisible) for the visible light because it is of high energy and gets easily absorbed in order to dislodge electrons. The infrared light carries low energy and does not pass easily through the photosphere; therefore, they are not opaque.
Explanation of Solution
Introduction:
The light that gets absorbed easily through a medium becomes opaque (invisible), but it remains capable enough to excite electrons in the material it passes through. The light that fails to excite electrons from a medium cannot pass through the same.
Explanation:
The energy carried by the infrared radiations is far less than that carried by the ultraviolet and/or the visible light. Higher energy is able to excite the electrons through the photosphere, which is primarily made up of negative hydrogen ions. As the visible light excites electrons, it gets easily absorbed in the photosphere, thus becomes invisible or opaque.
The infrared light of longer wavelength (or less energy) than the one calculated in part (a) will fail to excite the electrons from the photosphere. They will not pass through the photosphere, thus will become visible or opaque to naked eyes.
Conclusion:
Therefore, the infrared light is opaque to the photosphere while visible light is not because the former is of less energy and does not pass through the photosphere while the latter passes through and excites the electrons so as to become invisible.
Want to see more full solutions like this?
Chapter 16 Solutions
Universe
- When current is flowing through the coil, the direction of the torque can be thought of in two ways. Either as the result of the forces on current carrying wires, or as a magnetic dipole moment trying to line up with an external field (e.g. like a compass). Note: the magnetic moment of a coil points in the direction of the coil's magnetic field at the center of the coil. d) Forces: We can consider the left-most piece of the loop (labeled ○) as a short segment of straight wire carrying current directly out of the page at us. Similarly, we can consider the right-most piece of the loop (labeled ) as a short segment straight wire carrying current directly into the page, away from us. Add to the picture below the two forces due to the external magnetic field acting on these two segments. Then describe how these two forces give a torque and determine if the torque acts to rotate the loop clockwise or counterclockwise according to this picture? Barrow_forwardIn each of the following, solve the problem stated. Express your answers in three significant figures. No unit is considered incorrect. 1. For the circuit shown, determine all the currents in each branch using Kirchhoff's Laws. (3 points) 6 5V 2 B C 4 A www 6 VT ww T10 V F E 2. Compute for the total power dissipation of the circuit in previous item. (1 point) 3. Use Maxwell's Mesh to find Ix and VAB for the circuit shown. (3 points) Ix 50 V 20 ww 21x B 4. Calculate all the currents in each branch using Maxwell's Mesh for the circuit shown. (3 points) www 5ი 10 24V 2A 2002 36Varrow_forwardIf the mass of substance (1 kg), initial temperature (125˚C), the final temperature (175˚C) and the total volume of a closed container (1 m3) remains constant in two experiments, but one experiment is done with water ( ) and the other is done with nitrogen ( ). What is the difference in the change in pressure between water and nitrogen?arrow_forward
- Using the simplified energy balance in Equation 1, suppose there is heat transfer of 40.00 J to a system, while the system does 10.00 J of work. Later, there is heat transfer of 25.00 J out of the system while 4.00 J of work is done on the system. What is the net change in internal energy of the system?arrow_forwardYou pour a litre (1 kg) of 25.0˚C water into a 0.500 kg aluminium pan off the stove, but has previously been heated so it starts with a temperature of 120˚C. What is the temperature when the water and the pan reach thermal equilibrium (i.e., what is the temperature of both objects when they reach the same temperature)? Assume that the pan is placed on an insulated pad and a negligible amount of water boils off.arrow_forwardA golf club hits a golf ball and the golf ball’s flight reaches a maximum height of 5.48 m. Calculate the momentum of the golf ball at the maximum height if the mass of the golf ball is 0.459 kg.arrow_forward
- • Superposition Theorem • Thevenin's and Norton's Theorem 1. Find the unknown voltage V₁, unknown resistances R1 and R2, and currents flowing through R1 and R2 for the circuit shown below using Superposition Theorem. 40 V + R₁₂ w B C ♥16A 10A www 4A F ww 2 E Ꭰ 2. Use Thevenin's Theorem to find the current flowing in 3-ohm resistor and its power dissipation from the circuit shown in the right. + 3. Use Norton's Theorem for the same instruction as for No. 2. 8 V A www 202 B wwww 20 Ω 10 V + 302 202 www C - 12 V 502 www.arrow_forwardFill in blanksarrow_forwardA rock is dropped from a height of 2.00 m. Determine the velocity of the rock just before it hits the ground. If the momentum of the rock just before hitting the ground is 14.0 kg m/s, what is the mass of the rock? Is the collision between the rock and the ground elastic or inelastic? Explain.arrow_forward
- Describe how the momentum of a single ball changes as it free falls from a height of approximately 1 m, collides with a hard floor, and rebounds.arrow_forward• Nature of Resistance Temperature-Resistance Relationship Ohm's Law, Energy and Power Kirchhoff's Law • • Maxwell's Mesh Analysis 1. The steel of the third rail of a railway system has a resistivity of 21.4 μ-cm. If its cross-sectional area is 8.2 in², calculate the resistance per mile of rail, neglecting the effect of joints between sections. (1 point) 2. An incandescent lamp has a tungsten filament whose resistance is 96 at its operating temperature of 2900°C. Calculate the filament resistance when the lamp is disconnected from the electric source, under which condition its temperature is 24°C. (Use do = 0.0045 02/°C for tungsten) (1 point) 3. For the circuit shown, find the following: 50 V 602 10 V 702 a. the value of resistor R. (1 point) b. the equivalent resistance with respect to the 50-V source. (1 point) 4. For the circuit shown, determine all the currents in each branch using Kirchhoff's Laws. (3 points) A 5V 2 В -ний C 4 6 VT ww F E 5. Use Maxwell's Mesh to find I, and VAB…arrow_forwardFor items 8-9, refer to the problem below. Find all the currents flowing in every resistor, power dissipation in every resistor and the total power of the circuit shown at the right using... 8. Kirchhoff's Laws (5 pts) 9. Maxwell's Mesh Analysis (5 pts) A 8 V 10 V B + 20 Ω 3Ω 202 wwww C wwww 202 + 50 www 12 Varrow_forward
- Foundations of Astronomy (MindTap Course List)PhysicsISBN:9781337399920Author:Michael A. Seeds, Dana BackmanPublisher:Cengage LearningModern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage Learning
- Stars and Galaxies (MindTap Course List)PhysicsISBN:9781337399944Author:Michael A. SeedsPublisher:Cengage LearningUniversity Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStaxGlencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-Hill





