Explanation Given data: Mass of potassium atom 1 is 6.47 × 10 − 26 kg . Speed of potassium atom 1 is 1.3 × 10 4 m s . Speed of potassium atom 2 is 1.17 × 10 4 m s . Energy of released in forms of photon is 3.06 eV. Formula used: Consider the expression of energy. K i = K f + E light (1) Here, K i is the initial kinetic energy, K f is the final kinetic energy, and E light is the amount of energy released collision. Consider the expression of kinetic energy. K = 1 2 m v 2 (2) Here, m is the mass, and v is the velocity of the particle. The final kinetic energy after collision is the final kinetic energy of atom 1 and kinetic energy of atom 2. As the atom 2 is starting from rest the initial velocity of atom 2 is zero means its initial kinetic energy is zero. Modify equation (1) with respect to equation (2) 1 2 m ( v 1 ) initial 2 + 1 2 m ( v 2 ) initial 2 = 1 2 m ( v 1 ) final 2 + 1 2 m ( v 2 ) final 2 + E light ( v 1 ) initial 2 + ( v 2 ) initial 2 = ( v 1 ) final 2 + ( v 2 ) final 2 + ( E light ) 2 m Here, ( v 1 ) initial is initial velocity of atom 1, ( v 1 ) final is final velocity of atom 1, and ( v 2 ) final is final velocity of atom 2. Substitute 1.3 × 10 4 m s for ( v 1 ) initial , 1.17 × 10 4 m s for ( v 2 ) final , 0 for ( v 2 ) initial , 6.47 × 10 − 26 kg for m , and 3.06 eV for E light , ( 1.3 × 10 4 m s ) 2 + 0 = ( 1.17 × 10 4 m s ) 2 + ( v 1 ) final 2 + ( 3

Student Workbook for College Physics: A Strategic Approach Volume 2 (Chs. 17-30)

3rd Edition

ISBN: 9780321908872

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

See similar textbooks

Concept explainers

Atomic Spectra

According to the Bohr model of an atom, the electron in an atom moves around a nucleus in fixed orbits with specific energies known as energy levels. The orbital energies are quantized. The electrons remain in these energy levels until they emit or absor…

Ruby Lasers

Lasers are devices that emit light using atoms or molecules at a certain wavelength and amplify the light to produce a narrow beam of radiation. It works as per the principle of electromagnetic radiation. Their source of emission contains the same freque…

Balmer Series

The spectrum of frequency observed when electromagnetic radiation is emitted from an atom when it goes from higher energy state to lower state, is known as emission spectrum. This transition occurs when an excited electron moves from higher to lower stat…

Emission Spectrum

Every state of matter tries to be at minimum potential energy or it can be said that the atoms of element/ substance arrange themselves such that overall energy is minimum.

Question

Chapter 29, Problem 62GP

To determine

The final speed of atom 1.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 29, Problem 61GP

Chapter 29, Problem 63GP

Students have asked these similar questions

What are the (a) energy, (b) magnitude of the momentum, and (c) wavelength of the photon emitted when a hydrogen atom undergoes a transition from a state with n = 3 to a state with n = 1?

A sodium atom emits a photon with wavelength 818 nm shortly after being struck by an electron. What minimum speed did the electron have before the collision? Express your answer with the appropriate units Note the following answers are wrong 1.1 * 10^6 m/s this one is close because it said "not quite you may have made a rounding error" 1.16 * 10^6 m/s this one is close because it said "not quite you may have made a rounding error" 1 * 10^6 m/s 1.2 * 10^6 m/s

An electron in hydrogen absorbs a photon and jumps from orbit n = 2 to n = 4. Using the energy level diagram shown, what was the energy of the absorbed photon? a. 3.4 eV b. 10.2 eV c. 0.85 eV d. 0 eV e. 2.55 eV

Chapter 29 Solutions

Student Workbook for College Physics: A Strategic Approach Volume 2 (Chs. 17-30)

Ch. 29 - Prob. 1CQ Ch. 29 - Prob. 2CQ Ch. 29 - Prob. 3CQ Ch. 29 - Prob. 4CQ Ch. 29 - Prob. 5CQ Ch. 29 - Prob. 6CQ Ch. 29 - Prob. 7CQ Ch. 29 - Prob. 8CQ Ch. 29 - Prob. 9CQ Ch. 29 - Prob. 10CQ

Ch. 29 - Prob. 11CQ Ch. 29 - Prob. 12CQ Ch. 29 - Prob. 13CQ Ch. 29 - Prob. 14CQ Ch. 29 - Prob. 15CQ Ch. 29 - Prob. 16CQ Ch. 29 - Prob. 17CQ Ch. 29 - Prob. 18CQ Ch. 29 - Prob. 19CQ Ch. 29 - Prob. 20CQ Ch. 29 - Prob. 21CQ Ch. 29 - Prob. 22CQ Ch. 29 - Prob. 23CQ Ch. 29 - Prob. 24MCQ Ch. 29 - Prob. 25MCQ Ch. 29 - Prob. 26MCQ Ch. 29 - Prob. 27MCQ Ch. 29 - Prob. 28MCQ Ch. 29 - Prob. 29MCQ Ch. 29 - Prob. 30MCQ Ch. 29 - Prob. 1P Ch. 29 - Prob. 2P Ch. 29 - Prob. 3P Ch. 29 - Prob. 4P Ch. 29 - Prob. 5P Ch. 29 - Prob. 6P Ch. 29 - Prob. 7P Ch. 29 - Prob. 8P Ch. 29 - Prob. 9P Ch. 29 - Prob. 10P Ch. 29 - Prob. 11P Ch. 29 - Prob. 12P Ch. 29 - Prob. 13P Ch. 29 - Prob. 14P Ch. 29 - Prob. 15P Ch. 29 - Prob. 16P Ch. 29 - Prob. 17P Ch. 29 - Prob. 18P Ch. 29 - Prob. 19P Ch. 29 - Prob. 21P Ch. 29 - Prob. 22P Ch. 29 - Prob. 23P Ch. 29 - Prob. 24P Ch. 29 - Prob. 25P Ch. 29 - Prob. 26P Ch. 29 - Prob. 27P Ch. 29 - Prob. 28P Ch. 29 - Prob. 29P Ch. 29 - Prob. 30P Ch. 29 - Prob. 31P Ch. 29 - Prob. 32P Ch. 29 - Prob. 33P Ch. 29 - Prob. 34P Ch. 29 - Prob. 35P Ch. 29 - Prob. 36P Ch. 29 - Prob. 37P Ch. 29 - Prob. 38P Ch. 29 - Prob. 39P Ch. 29 - Prob. 40P Ch. 29 - Prob. 41P Ch. 29 - Prob. 42P Ch. 29 - Prob. 43P Ch. 29 - Prob. 44GP Ch. 29 - Prob. 45GP Ch. 29 - Prob. 46GP Ch. 29 - Prob. 47GP Ch. 29 - Prob. 48GP Ch. 29 - Prob. 49GP Ch. 29 - Prob. 50GP Ch. 29 - Prob. 51GP Ch. 29 - Prob. 52GP Ch. 29 - Prob. 53GP Ch. 29 - Prob. 54GP Ch. 29 - Prob. 55GP Ch. 29 - Prob. 56GP Ch. 29 - Prob. 57GP Ch. 29 - Prob. 58GP Ch. 29 - Prob. 59GP Ch. 29 - Prob. 60GP Ch. 29 - Prob. 61GP Ch. 29 - Prob. 62GP Ch. 29 - Prob. 63GP Ch. 29 - Prob. 64GP Ch. 29 - Prob. 65GP Ch. 29 - Prob. 66GP Ch. 29 - Prob. 67GP Ch. 29 - Prob. 68GP Ch. 29 - Prob. 69GP Ch. 29 - Prob. 70GP Ch. 29 - Prob. 71GP Ch. 29 - Prob. 73GP Ch. 29 - Prob. 74GP Ch. 29 - Prob. 75GP Ch. 29 - Prob. 76MSPP Ch. 29 - Prob. 77MSPP Ch. 29 - Prob. 78MSPP Ch. 29 - Prob. 79MSPP

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

The final speed of atom 1.

Solution Summary: The author explains that the final speed of atom 1 is 4120, while the energy of released in forms of photon is 3.06 eV.

Concept explainers

The final speed of atom 1.

Want to see the full answer?

Chapter 29 Solutions