Physics for Scientists and Engineers with Modern Physics
4th Edition
ISBN: 9780131495081
Author: Douglas C. Giancoli
Publisher: Addison-Wesley
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Chapter 39, Problem 25P
To determine
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(a) How much energy is required to cause an electron in hydrogen to move from the n = 2 state to the n = 5 state?
in J(b) Suppose the atom gains this energy through collisions among hydrogen atoms at a high temperature. At what temperature would the average atomic kinetic energy 3/2 * kBT be great enough to excite the electron? Here kB is Boltzmann's constant.
in K
(a) How much energy is required to cause an electron in hydrogen to move from the n = 2 state to the n = 5 state?in J(b) Suppose the atom gains this energy through collisions among hydrogen atoms at a high temperature. At what temperature would the average atomic kinetic energy 3/2 * kBT be great enough to excite the electron? Here kB is Boltzmann's constant.
in K
(a)
The Lyman series in hydrogen is the transition from energy levels n = 2, 3, 4, ...
to the ground state n =
1. The energy levels are given by
13.60 eV
En
n-
(i)
What is the second longest wavelength in nm of the Lyman series?
(ii)
What is the series limit of the Lyman series?
[1 eV = 1.602 x 1019 J, h = 6.626 × 10-34 J.s, c = 3 × 10° m.s]
%3D
Two emission lines have wavelengts A and + A2, respectively, where AA <<2.
Show that the angular separation A0 in a grating spectrometer is given
aproximately by
(b)
A0 =
V(d/m)-2
where d is the grating constant and m is the order at which the lines are observed.
Chapter 39 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 39.2 - Prob. 1AECh. 39.2 - Prob. 1BECh. 39.3 - Prob. 1CECh. 39.4 - Prob. 1DECh. 39.4 - Prob. 1EECh. 39.5 - Prob. 1FECh. 39.7 - Prob. 1GECh. 39 - Prob. 1QCh. 39 - Prob. 2QCh. 39 - Prob. 3Q
Ch. 39 - Prob. 4QCh. 39 - Prob. 5QCh. 39 - Prob. 6QCh. 39 - Prob. 7QCh. 39 - Prob. 8QCh. 39 - Prob. 9QCh. 39 - Prob. 10QCh. 39 - Prob. 11QCh. 39 - On what factors does the periodicity of the...Ch. 39 - Prob. 13QCh. 39 - Prob. 14QCh. 39 - Prob. 15QCh. 39 - Prob. 16QCh. 39 - Prob. 17QCh. 39 - Prob. 18QCh. 39 - Prob. 19QCh. 39 - Prob. 20QCh. 39 - Prob. 21QCh. 39 - Prob. 22QCh. 39 - Prob. 23QCh. 39 - Prob. 24QCh. 39 - Prob. 25QCh. 39 - Prob. 26QCh. 39 - Prob. 27QCh. 39 - Prob. 28QCh. 39 - Prob. 29QCh. 39 - Prob. 1PCh. 39 - Prob. 2PCh. 39 - Prob. 3PCh. 39 - Prob. 4PCh. 39 - Prob. 5PCh. 39 - Prob. 6PCh. 39 - Prob. 7PCh. 39 - Prob. 8PCh. 39 - Prob. 9PCh. 39 - Prob. 10PCh. 39 - Prob. 11PCh. 39 - Prob. 12PCh. 39 - Prob. 13PCh. 39 - Prob. 14PCh. 39 - Prob. 15PCh. 39 - Prob. 16PCh. 39 - Prob. 17PCh. 39 - Prob. 18PCh. 39 - Prob. 19PCh. 39 - Prob. 20PCh. 39 - Prob. 21PCh. 39 - Prob. 22PCh. 39 - Prob. 23PCh. 39 - Prob. 24PCh. 39 - Prob. 25PCh. 39 - Prob. 26PCh. 39 - Prob. 27PCh. 39 - Prob. 28PCh. 39 - Prob. 29PCh. 39 - Prob. 30PCh. 39 - Prob. 31PCh. 39 - Prob. 32PCh. 39 - Prob. 33PCh. 39 - Prob. 34PCh. 39 - Prob. 35PCh. 39 - Prob. 36PCh. 39 - Prob. 37PCh. 39 - Prob. 38PCh. 39 - Prob. 39PCh. 39 - Prob. 40PCh. 39 - Prob. 41PCh. 39 - Prob. 42PCh. 39 - Prob. 43PCh. 39 - Prob. 44PCh. 39 - Prob. 45PCh. 39 - Prob. 46PCh. 39 - Prob. 47PCh. 39 - Prob. 48PCh. 39 - Prob. 49PCh. 39 - Prob. 50PCh. 39 - Prob. 51PCh. 39 - Prob. 52PCh. 39 - Prob. 53PCh. 39 - Prob. 54PCh. 39 - Prob. 55PCh. 39 - Prob. 56PCh. 39 - Prob. 57PCh. 39 - Prob. 58PCh. 39 - Prob. 59PCh. 39 - Prob. 60PCh. 39 - Prob. 61GPCh. 39 - Prob. 62GPCh. 39 - Prob. 63GPCh. 39 - Prob. 64GPCh. 39 - Prob. 65GPCh. 39 - Prob. 66GPCh. 39 - Prob. 67GPCh. 39 - Prob. 68GPCh. 39 - Prob. 69GPCh. 39 - Prob. 70GPCh. 39 - Prob. 71GPCh. 39 - Prob. 72GPCh. 39 - Prob. 73GPCh. 39 - Prob. 74GPCh. 39 - Prob. 75GPCh. 39 - Prob. 76GPCh. 39 - Prob. 77GP
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- .55 The radial probability density for the ground state of the hydrogen atom is a maximum whenr = a, where a is the Bohr ra- dius. Show that the average value of r, defined as P(r) r dr, has the value 1.5a. In this expression for ravgs each value of P(r) is weighted with the value of r at which it occurs. Note that the average value of r is greater than the value of r for which P(r) is a maximum.arrow_forwardThe equation for the energy of a K« x – ray is Exa = (10.2 eV)(Z – 1)2. Explain where the 10.2 eV comes from and why the effective value of Z is Z – 1.arrow_forward(a) Show that the speed of an electron in the nth Bohr orbit of hydrogen is ac/n, where a is the fine structure constant, equal to e/4neghc. (b) What would be the speed in a hydrogen like atom with a nuclear charge of Ze?arrow_forward
- (ii):Find the ground state L and S of oxygen (Z=8). Also find the matter wave associated with 111 MeV α-particles. Moreover, using L-S coupling find the ground state term for nd8 .arrow_forwardIf elements beyond Z = 120 are ever synthesized, electrons in these heavy atoms will begin filling a g subshell, corresponding to l = 4. How many states will be in a g subshell?arrow_forwardwhere ?∞ = 1.097 × 10^7 m−1is the Rydberg constant and ? is the atomic number (thenumber of protons found in the nucleus). Calculate the ground state energy of a triplyionised beryllium atom, Be3+ (a beryllium atom with three electrons removed).arrow_forward
- (4) Electronic energy level of a hydrogen atom is given by R E ; n = n2 1, 2, 3,... and R = 13.6 eV. Each energy level has degeneracy 2n² (degeneracy is the number of equivalent configurations associated with the energy level). (a) Calculate the partition function Z for a hydrogen atom at a constant temperature. (b) Let us consider that the energy level of a hydrogen atom is approximated by a two level system, n = 1,2. Estimate the mean energy at 300 K.arrow_forward(a) The L→ K transition of an X-ray tube containing a molybdenum (Z = 42) target occurs at a wavelength of 0.0724 nm. Use this information to estimate the screening parameter of the K-shell electrons in molybdenum. [Osmania University]arrow_forwardCheck Your Understanding Can the magnitude of Lzever be equal to L?arrow_forward
- Determine the integral | P(r) dr for the radial probability density for the ground state of the hydrogen atom 4 P(r) = - r²e-2rla a³ O 1 O-1 O 0.5arrow_forwardShow that at limit ħo kgT, the following expression, h €(w, T) = c exp(Bho)-1 reduces to the classical form given by: e(w, T)= Ac ³(kgT)w²: ² = 0 [₁²³ (²+²)]. Acarrow_forwardAssume that the 1+z) and |-z) states for an electron in a magnetic field are energy eigen- vectors with energies E and 0, respectively, and assume that the electron's state at t = 0 is -M |y(0)) = Find the probability that we will determine this electron's spin to be in the +x direction at time tarrow_forward
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