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ANTONII [103]
3 years ago
14

In a photoelectric experiment it is found that a stopping potential of 1.00 V is needed to stop all the electrons when incident

light of wavelength 278 nm is used and 2.6 V is needed for light of wavelength 207 nm. From these data determine Planck's constant and the work function of the metal.
Physics
1 answer:
kiruha [24]3 years ago
5 0

Answer:

Planck’s constant is 6.9*10^{-34} Js

work function of metal is 5.846*10^{-19}

Explanation:

The maximum kinetic energy from electron from photoelectric effect is given by

\K_{max}=eV_o where V_o  is applied voltage and e is charge on electron and substituting charge on electron by 1.6*10^{-19}  and 1.00 V for V_o  

K_{max}=1.6*10^{-19}*1=1.6*10^{-19} J

Considering that the wavelength, \lambda of light used is given as 278*10^{-9} m

Energy of light is given by

E=\frac {hc}{ \lambda}   where \lambda  is the wavelength, h is Planck’s constant and c is speed of light. Taking c for 3*10^{8} m/s  

Substituting values of wavelength and speed of light we obtain

E=\frac {h*3*10^{8}}{278*10^{-9}}=h1.07914*10^{15} J

If work function is \phi  then

E=\phi + k_{max}  hence

h1.07914*10^{15} J=\phi +1.6*10^{-19} J ------- Equation 1

Energy corresponding to wavelength of 207 nm is

E=h\frac {3*10^{8} m/s}{207*10^{-9}}=h(1.45*10^{15}}) J

Maximum kinetic energy of electrons when V_o  is 2.6 V becomes

K_{max}=(1.6*10^{-19})*2.6V=4.16*10^{-19} J

From E=\phi + k_{max}  and substituting h(1.45*10^{15}}) J  for E and 4.16*10^{-19} J  for K_{max}  we have

h(1.45*10^{15}}) J=\phi +4.16*10^{-19} J ------ Equation 2

Equation 2 minus equation 1 gives

h3.71*10^{14}=2.56*10^{-19}

h=\frac {2.56*10^{-19}}{3.71*10^{14}}\approx 6.9*10^{-34} Js

Therefore, Planck’s constant is 6.9*10^{-34} Js

Recalling equation 1 and substituting back the value of h as obtained

h1.07914*10^{15} J=\phi +1.6*10^{-19} J

(6.9*10^{-34})*(1.07914*10^{15})=\phi +1.6*10^{-19} J

\ph=(6.9*10^{-34})*(1.07914*10^{15})-(1.6*10^{-19})=5.846*10^{-19}

Therefore, work function of metal is 5.846*10^{-19}

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A 2137 kg car moving east at 12.91 m/s collides with a 3264 kg car moving north. The cars stick together and move as a unit afte
IRINA_888 [86]

Answer:

The speed of the 3264 kg car before collision is 6.32 m/s

Explanation:

Let first car, mass m₁, moving east be moving with velocity, v₁ and second car, mass m₂, moving north be moving with velocity, v₂. Let the velocity of the two cars after collision be v₃

In a Collison (whether elastic or inelastic), momentum is always conserved.

Momentum before collision = momentum after collision.

Momentum before collision = m₁v₁ + m₂v₂

v₁ = (12.91î) m/s in vector form, m₁ = 2137 kg

v₂ = ?, v₂ = (v₂j) m/s in vector form, m₂ = 3264 kg

Momentum before collision = (2137)(12.91î) + (3264)(v₂j) = (27590î + 3264v₂j) kgm/s

Momentum after collision = (total mass of the cars after collision) × (velocity of the stuck-together cars after collision)

Total mass of the cars after collision = m₁ + m₂ = 2137 + 3264 = 5401 kg

Velocity after collision, v₃ = 6.38 m/s In the N36.8°E direction, put in vector form,

v₃ = (6.38 cos 36.8°î + 6.38 sin 36.8°j)

v₃ = (5.11î + 3.82j) m/s

Momentum after collision = 5401 (5.11î + 3.82j) = (27590î + 20641.4j) kgm/s

Momentum before collision = momentum after collision.

(27590î + 3264v₂j) = (27590î + 20641.4j)

Comparing components

3264v₂ = 20641.4j

v₂ = 6.32 m/s

Hope this Helps!!!

6 0
3 years ago
How to measure the speed of sound in air?
Leona [35]

STEP ONE:


Let  you and your friend stand as far away as possible from a large reflecting wall and clap your hands rapidly at a regular rate.  


STEP TWO:


Adjust this rate until each clap just coincides with the return of an echo of its predecessor, or until clap and echo are heard as equally spaced.


STEP THREE:


Use a stopwatch to find the time between claps, t. Make a rough measurement of distance to the wall, s. Thus the speed of sound, v = 2s/t

8 0
3 years ago
Followed by the previous question: presume that the electron performs a uniform circular motion around the hydrogen nucleus. Wha
Ksivusya [100]

Answer:

A_c=87.73*10^{21}m/s

Explanation:

From the question we are told that

r=5\times 10^{-11}

T=1.5 \times 10^{-16}

Generally the equation for velocity is mathematically given as

Velocity (v)=\frac{2 \pi r}{t}

V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}

V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}

Generally the equation for Centripetal acceleration is mathematically given as

A_c=\frac{V^2}{r}

A_c=(\frac{20.944*10^5)}{r5*10^{-11}}

A_c=87.73*10^{21}m/s

8 0
2 years ago
What is the change in velocity of a 22-kg object that experiences a force of 15 N for
vagabundo [1.1K]

Answer:

Force = mass × acceleration

Acceleration:

{ \tt{15 = (22 \times a)}} \\ { \tt{a =  \frac{15}{22}  \:  {ms}^{ - 2} }}

From first Newton's equation of motion:

{ \bf{v = u + at}}

Change = v - u:

{ \tt{v - u = (a \times t)}} \\ { \tt{v - u = ( \frac{15}{22} \times 1.2) }} \\ { \tt{v - u = 0.82 \:  {ms}^{ - 2} }}

3 0
3 years ago
An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F
ololo11 [35]

Answer

given,

mass of jogger  = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed  = 1.3 m/s  in  61 ° north of east.

jogger one

P_1 = m_1 v_1 \hat{i}

P_1 = 67 \times 2.3\hat{i}

P_1 = 154.1 \hat{i}

P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}

P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}

P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}

P_2 = 44.12\hat{i} +79.59\hat{j}

now

P = P₁ + P₂

P = 198.22 \hat{i} +79.59 \hat{j}

magnitude

P = \sqrt{198.22^2 + 79.59^2}

P =213.60 kg.m/s

\theta = tan^{-1}\dfrac{79.59}{198.22}

\theta = 21.87

the angle is \theta = 21.87 north of east

7 0
3 years ago
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