What is the nergy of a wave that has a frequency of 6.32 x 10^20 Hz?

Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

Nuetrik [128]

Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(c) $A = 5.51 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

(c) For face centered cubic lattice:

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

6 0

3 years ago

Calculate the binding energy per nucleon of the helium nucleus 52he. express your answer in millions of electron volts to four s

Vitek1552 [10]

The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for 52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=binding energy
52He-----------> 2 x 11p + 3 x10n is the equation bilan
so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
mn=mass of neutron=1.67 10^-27 kg
 m nucleus= 5
Dm= 2x1.67 10^-27 kg+ 3x1.67 10^-27 kg-5= - 4.9 J
Eb= - 4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so -9.10^16 J/ 1.6 10^-19= -5.625 10^35 eV

the final answer is Eb /nucleon = -5.625 x10^35 eV

8 0

3 years ago

A vertical wire carries current in the upward direction. an electron is traveling parallel to the wire. what is the angle α betw

Luden [163]

Angle α between the velocity of theelectron and the magnetic field of the wire will be 90 degree

7 0

3 years ago

A cannon is fired horizontally at 243 m/s off of a 62 meter tall, shear vertical cliff. How far in meters from the base of the c

Alekssandra [29.7K]

Answer:

865.08 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 243 m/s

Height (h) of the cliff = 62 m

Horizontal distance (s) =?

Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:

Height (h) of the cliff = 62 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

62 = ½ × 9.8 × t²

62 = 4.9 × t²

Divide both side by 4.9

t² = 62/4.9

Take the square root of both side.

t = √(62/4.9)

t = 3.56 s

Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:

Initial velocity (u) = 243 m/s

Time (t) = 3.56 s

Horizontal distance (s) =?

s = ut

s = 243 × 3.56 s

s = 865.08 m

Thus, the cannon ball will impact the ground 865.08 m from the base of the cliff.

6 0

3 years ago

Iii A load of 2.5 N is hung on the spring.

Anna11 [10]

N stands for Newton
2- upload the graph..

7 0

3 years ago