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3 years ago

9

The energy required to dislodge electrons from sodium metal via the photoelectric effect is 275 kj/mol. What wavelength in nm of

light has sufficient energy per photon to dislodge an electron from the surface of sodium?

1 answer:

denis23 [38]3 years ago

7 0

Answer: 430 nm

Explanation:

For 257 kJ to dislodge one mole of electrons we need,

275 x 10^3 / 6 x 10^23 = 4.6 x 10^-19

Using Einstein’s relationship between energy, frequency and wavelength

E = hf = h x (c/landa)

Therefore

Landa = h x (c/E) = 6.6 x 10^-34 x (3 x 10^8 / 4.6 x 10^-19 = 4.3 x 10-7 m.

In nm, landa = 430 nm

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Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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3 years ago

In a game of pool, a cue ball rolls without slipping toward the stationary eight ball with a momentum of 0.23 kg. After the two

IrinaVladis [17]

This question involves the concepts of the law of conservation of momentum.

The magnitude of the final momentum of the eight ball is "0.22 N.s".

According to the law of conservation of momentum:

$P_{i1}+P_{i2}=P_{f1}+P_{f2}$

where,

$P_{i1}$ = initial momentum of the cue ball = 0.23 N.s

$P_{i2}$ = initial momentum of the eight ball = 0 N.s (since ball is initially at rest)

$P_{f1}$ = final momentum of the cue ball = 0.01 N.s

$P_{f2}$ = final momentum of the eight ball = ?

Therefore,

$0.23\ N.s + 0\N.s = 0.01\ N.s+P_{f2}\\\\P_{f2} = 0.22\ N.s$

Learn more about the law of conservation of momentum here:

brainly.com/question/1113396?referrer=searchResults

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3 years ago

An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and

Svetradugi [14.3K]

Answer:

Total contraction on the Bar = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm

Diameter for aluminum bar = 40 mm

Hole diameter = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180 × 10³ N

Calculate the total contraction on the bar = ???

The relation used in calculating the contraction on the bar is:

$\delta L = \dfrac{P *L }{A*E}$

The relation used in calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Total contraction on the Bar = $\dfrac{180 *10^3 \N }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]$

Total contraction on the Bar = 2.117( 0.398 + 0.182)

Total contraction on the Bar = 2.117*(0.58)

Total contraction on the Bar = 1.22786 mm

5 0

3 years ago