An electron has de Broglie wavelength 2.78×10-10 m . Part A Determine the magnitude of the electron's momentum pe. Express your answer in kilogram meters per second to three significant figures. ▸ View Available Hint(s) VG ΑΣΦ 24 Pe 2.29 10 . Submit Previous Answers ? × Incorrect; Try Again; One attempt remaining kg. m/s
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- A neutron of mass 1.675 × 10-27 kg has a de Broglie wavelength of 7.8x10-12 m. What is the kinetic energy (in eV) of this non-relativistic neutron? Please give your answer with two decimal places. 1 eV = 1.60 × 10-19 J, h = 6.626 × 10-34 J ∙ s.For the thermal radiation from an ideal blackbody radiator with a surface temperature of 2000 K, let Ic represent the intensity per unit wavelength according to the classical expression for the spectral radiancy and IP represent the corresponding intensity per unit wavelength according to the Planck expression.What is the ratio Ic/IP for a wavelength of (a) 400 nm (at the blue end of the visible spectrum) and (b) 200 mm (in the far infrared)? (c) Does the classical expression agree with the Planck expression in the shorter wavelength range or the longer wavelength range?What is the de Broglie wavelength for an electron with speed (a) v = 0.469c and (b) v = 0.958c? (Hint: Use the correct relativistic expression for linear momentum if necessary.)
- (PART 1)Light of wavelength 350 nm falls on a potassium surface, and the photoelectrons have amaximum kinetic energy of 1.3 eV.What is the work function of potassium?The speed of light is 3 × 108 m/s and Planck’sconstant is 6.63 × 10−34 J · s.Answer in units of eV (PART 2) What is the threshold frequency for potassium?Answer in units of Hz.A 279 lb fullback runs the 40 yd dash at a speed of 18.5 ×× 0.1 mi/h. (a) What is his de Broglie wavelength (in meters)? (b) What is the uncertainty of his position?Light of wavelength 350 nm falls on a potassium surface, and the photoelectrons have amaximum kinetic energy of 1.3 eV.What is the work function of potassium?The speed of light is 3 × 108 m/s and Planck’sconstant is 6.63 × 10−34 J · s.Answer in units of eV. What is the threshold frequency for potassium?Answer in units of Hz.
- UV radiation having a wavelength of 84 nm falls on gold metal, to which electrons are bound by 4.82 eV. What is the maximum velocity of the ejected photoelectrons? No need to use relativistic formulas in this case, so you can just use the standard formula KE =12mv2. The correct answer is 1.87E6 m/s how do I get that?The quantum-mechanical treatment of the hydrogen atomgives the energy, E, of the electron as a function of the principal quantum number, n:E=-h²/8π²mₑ²aₒ²n² (n=123...) where his Planck’s constant, meis the electron mass, and a0 is 52.92X10⁻¹² =-m.(a) Write the expression in the form E(constant)=-(constant)1/n² , evalu-ate the constant (in J), and compare it with the corresponding expression from Bohr’s theory.(b) Use the expression to find ΔE between n=2 and n=3.(c) Calculate the wavelength of the photon that corresponds tothis energy change. Is this photon seen in the hydrogen spectrumobtained from experiment3. Using the average speed of a gas, (8RT?MW)1/2, determine the average de Broglie wavelength for an He atom at 25 °C and at 500 °C.How fast would the He atom need to travel in order to have the same linear momentum as a 500 nm photon?
- (b) Calculate the de Broglie wavelength of an electron having a mass of 9.11 x 10-31 kg and a charge of 1.602 x 10-19 J with a Kinetic energy of 110 eV. The value of the Planck’s constant is equal to 6.63 * 10-34 Js.What is the de Broglie wavelength of a electron that is moving at 8.77 x 105 m/s? Please give your answer in nanometers.Use the de Broglie relationship to determine the wavelengthsof the following objects: (a) an 85-kg person skiing at50 km>hr, (b) a 10.0-g bullet fired at 250 m>s, (c) a lithiumatom moving at 2.5 * 105 m>s, (d) an ozone 1O32 moleculein the upper atmosphere moving at 550 m>s.