Choose two of the qeustions and answer them with details please. n microscope em

Question

Choose two of the qeustions and answer them with details please.

n microscope employs a beam of electrons to 7.85 An electro obtain an image of an object. What energy must be imparted to each electron of the beam to obtain a wavelength of 10.0 pm? Obtain the energy in electron volts (eV) (1 eV = 1.602 × 1019 J). 7.86 Neutrons are used to obtain images of the hydrogen atoms in molecules. What a neutron beam to obtain a wavelength of 10.0 pm? Obtain the energy in electron volts (eV) (1 eV = 1.602 × 10-19 J). energy must be imparted to each neutron in 87 What is the number of different orbitals in each of the fol lowing subshells? 7.88 What is the number of different orbitals in each of the following subshells? List the possible subshells for the n = 6 shell. List the possible subshells for the n 7 shell. 7.89 7.90 7.91 What are gamma rays? How does the gamma radiation of improve their shelf life? foods i 7.92 How can gamma rays that are used in food irradiation be produced? Does such irradiated food show any radioactivity? 7.93 The word laser is an acronym meaning light amplification by stimulated emission of radiation. What is the stimulated emission of radiation? 7.94 Explain how lasers are used to "read" a compact disc 7.95 Explain the concept of quantum mechanical tunneling. 7.96 Explain how the probe in a scanning tunneling microscope scans a sample on the surface of a metal.

Explanation / Answer

Solution for Q 7.85

Given data

Wavelength = 10.0 pm

Energy = ?

Formula to calculate the energy using the wavelength is as follows

E= h*c/ wavelength (lambda)

Where

E= energy , h= planks constant (6.626E-34 J.s) and c= speed of light (3E8 m/s)

Lets convert wavelength from pm to meter

10.0 pm * 1 m / 1E12 pm = 1.0E-11 m

Now lets calculate the energy of electron

E =(6.626E-34 J.s * 3E8 m/s)/1.0E-11 m

E= 1.988E-14 J

Now lets convert energy from Joules to eV

1.988E-14 J * 1 eV / 1.602E-19 J = 1.24E5 eV

Therefore energy needed = 1.24E5 eV

Solution for the Q 7.87

Number of different orbital’s in each sub shell are follows

S sub shell only has 1 orbital

p sub shell has 3 different orbitals

d sub shell has 5 different orbitals

f sub shell has 7 different orbitals

therefore correct answers are as follows

a)3d = 5 orbitals

b)4f = 7 orbitals

c)4p = 3 orbitals

d)5s= 1 orbital

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.