Ask question

Ask question

All categories

2 years ago

6

The energy difference between the 5d and the 6s sublevels in gold accounts for its color. If this energy difference is about 2.7

eV (electron volt; 1 eV = 1.602 × 10−19 J), calculate the wavelength of light absorbed in the transition of an electron from the 5d subshell to the 6s subshell. Round the answer to the correct number of significant figures.

1 answer:

Anastaziya [24]2 years ago

8 0

Answer:

The wavelength of light absorbed in the transition is 459 nm.

Explanation:

Energy difference between 5-d and the 6-s sub-levels in gold = ΔE

$\Delta E=2.7 eV=2.7 eV\times 1.602\times 10^{-19} J=4.3254\times 10^{-19} J$

Let the wavelength of light absorbed in the transition 5-d to 6-s be $\lambda$

The relation between energy and wavelength is given by:

$E=\frac{h\times c}{\lambda}$

where,

E = energy of photon of the light

h = Planck's constant = $6.63\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of the photon

$4.3254\times 10^{-19} J=\frac{6.63\times 10^{-34}Js\times 3\times 10^8 m/s}{\Lambda }$

$\Lambda =4.59\times 10^{-7} m = 459 nm$

$1nm = 10^{-9 } nm$

The wavelength of light absorbed in the transition is 459 nm.

You might be interested in

Which of the following statements are true about an object in two-dimensional projectile motion with no air resistance? (There c

ki77a [65]

Answer:

The correct answers are

The following statements are true about an object in two-dimensional projectile motion with no air resistance

D) The speed of the object is zero at its highest point.

E) The horizontal acceleration is always zero and the vertical acceleration is always a non-zero constant downward

Explanation:

A) The speed of the object is constant but its velocity is not constant.

False the vertical velocity increases on descent

B) The acceleration of the object is constant but its object is + g when the object is rising and -g when it is falling.

False, the acceleration is -g when the object is rising

C) The acceleration of the object is zero at its highest point.

False, the acceleration is constant in magnitude throughout the motion

D) The speed of the object is zero at its highest point.

True, the direction of motion changes at the highest point from hence the body comes to rest and the speed is zero

E) The horizontal acceleration is always zero and the vertical acceleration is always a non-zero constant downward

True, the horizontal acceleration has associated force during motion but the vertical acceleration is due to gravity which is constant downwards

6 0

2 years ago

A Porsche challenges a Honda to a 200-m race.Because the Porsche's acceleration of 3.5 m/s2 is larger than the Honda's 3.0 m/s2,

Blizzard [7]

Answer:

Honda won by 0.14 s

Explanation:

We are given that

Distance =S=200 m

Initial velocity of Honda=u=0m/s

Initial velocity of Porsche=u'=0m/s

Acceleration of Honda= $3.0m/s^2$

Acceleration of Porsche's= $3.5m/s^2$

Time taken by Honda to start=1 s

$s=ut+\frac{1}{2}at^2$

Substitute the values

$200=0(t)+\frac{1}{2}(3)t^2$

$200=\frac{3}{2}t^2$

$t^2=\frac{200\times 2}{3}=\frac{400}{3}$

$t=\sqrt{\frac{400}{3}}=11.55s$

Time taken by Honda=11.55 s

Now, time taken by Porsche

$200=\frac{1}{2}(3.5)t^2$

$t^2=\frac{200\times 2}{3.5}$

$t=\sqrt{\frac{400}{3.5}}=10.69 s$

Total time taken by Porsche=10.69+1=11.69 s

Because it start 1 s late

Time taken by Honda is less than Porsche .Therefore, Honda won and

Time =11.69-11.55=0.14 s

Honda won by 0.14 s

3 0

2 years ago

Assume that a uniform magnetic field is directed into thispage. If an electron is released with an initial velocity directedfrom

Feliz [49]

Answer:

B). to the right

Explanation:

Since the direction of magnetic field is into the page

So here we know that

$B = B_o(-\hat k)$

now the velocity is from bottom to top

so we have

$v = v_o \hat j$

now the force on the moving charge is given as

$\vec F = q(\vec v \times \vec B)$

now we have

$\vec F = (-e)(v_o \hat j \times B_o(-\hat k))$

$\vec F = e v_o B \hat i$

so force will be towards Right

7 0

2 years ago

A common carnival ride, called a gravitron, is a large r = 11 m cylinder in which people stand against the outer wall. the cylin

Leya [2.2K]

Answer:

v = 13.19 m / s

Explanation:

This problem must be solved using Newton's second law, we create a reference system where the x-axis is perpendicular to the cylinder and the Y-axis is vertical

X axis

N = m a

Centripetal acceleration is

a = v² / r

Y Axis

fr -W = 0

fr = W

The force of friction is

fr = μ N

Let's calculate

μ (m v² / r) = mg

μ v² / r = g

v² = g r / μ

v = √ (g r /μ)

v = √ (9.8 11 / 0.62)

v = 13.19 m / s

7 0

2 years ago

A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin

KengaRu [80]

a) 219.8 rad/s

b) $20.0 rad/s^2$

c) $2.9 m/s^2$

d) $7005 m/s^2$

e) Towards the axis of rotation

f) $0 m/s^2$

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

$\omega_{max}=35 rev/s$

The angle covered during 1 revolution is

$\theta=2\pi$ rad

Therefore, the maximum angular velocity is:

$\omega_{max}=35 \cdot 2\pi = 219.8 rad/s$

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

$\alpha = \frac{\Delta \omega}{t}$

where

$\Delta \omega$ is the change in angular velocity

t is the time elapsed

Here we have:

$\omega_0 = 0$ is the initial angular velocity

$\omega_{max}=219.8 rad/s$ is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

$\alpha = \frac{219.8-0}{11}=20.0 rad/s^2$

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

Here we have

$\alpha =20.0 rad/s^2$

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

$a_t=(20.0)(0.145)=2.9 m/s^2$

d)

The magnitude of the radial (centripetal) acceleration is given by

$a_c = \omega^2 r$

where

$\omega$ is the angular velocity

r is the radius of the circle

Here we have:

$\omega_{max}=219.8 rad/s$ is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

$a_c=(219.8)^2(0.145)=7005 m/s^2$

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

$\alpha=0$ , since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

$a_t=(0)(0.145)=0 m/s^2$

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

$\omega=219.8 rad/s$

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

$v=(219.8)(0.145)=31.9 m/s$

7 0

2 years ago