# #1 Solutions

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## Q1.1

λmax T=2.9x10(-3)mK
λmax=145 nm=1.45x10(-7) m
T= (2.9x10(-3)mK)/(1.45x10(-7)m)= 2x104 K

## Q1.2

KE = hν - ϕ
0 = hν-2.24eV
2.24eV=(ν)x(4.14x10(-15) eV∙s)
ν= 5.4x1014 Hz
λ= c/ν= (3x108 m/s)/(5.4x1014 Hz)= 5.56x10(-7) m= 556 nm

## Q1.3

(a)

$KE=h\nu-\Phi$

$h\nu = KE +\Phi=49 eV + 4.7 eV=53.7 eV=53.7 eV \times 1.6 \times 10^{-19} \dfrac{J}{eV}=8.59\times 10^{-18} J$

$\nu=\dfrac{ 8.59\times 10^{-18} J }{h}= \dfrac{ 8.59\times 10^{-18} J }{6.63\times 10^{-34}J s} =1.30 \times 10^{16} s^{-1}$

$c=\lambda \nu$

So

$\lambda=\dfrac{c}{\nu}=\dfrac{3\times 10^{8}m/s}{1.30 \times 10^{16} /s}=2.32 \times 10^{-8} m=23.2 nm$

(b) Assuming every photon that hits the electrode ejects one electron, then we just need to find out how many photons will hit the electrode during the 2.0 second pulse.

The intensity of light is 10.0 W, so the energy of the light that shines on the electrode within 2.0 second period is: 10.0 W x 2 s=20.0 Joules.

The energy of each photon is $$8.59 \times 10^{-18} J$$, so the number of photons that hit the electrode within 2 seconds is:

$\dfrac{20.0 J}{8.59 \times 10^{-18} J}=2.33 \times 10^{18}$

This is also the number of electrons ejected within 2 seconds.

(c) The work function of the metal is $$\Phi=4.7 eV$$, so the minimum energy needed for a photon to eject an electron is when the kinetic energy of ejected electron is zero:

$KE=0=E_{photon}-\Phi$

$E_{photon}=\Phi=4.7 eV=4.7 eV \times 1.6 \times 10^{-19} \dfrac{J}{eV}=7.52 \times 10^{-19} J$

So when the energy of the photon is $$< 7 \times 10^{-19} J$$, then it can never eject an electron and there will be no electron ejected.

## Q1.4

2mW = 2mJ/s
a) Eph = hc/λ= 4.97x10(-19)J
Number of photons incident per second = number of electrons ejected per second = (2x10(-3)J)/4.97x10(-19)J =
4x1015el/s

b) KE = hν – ϕ
KE= 4.97x10(-19) J- 4.34x10(-19) J= 6.3x10(-20)J per electron
Total Power is equal to (KE)x(number of electrons per second) = 2.52x10(-4)W or 0.252m W

## Q1.5

1/λ=RH(1/12 - 1/n2 )= RH(1/1- 1/4)= 3RH/4
λ= 4/3RH = 4/(3x1.1x107 m)=1.2x10(-7)m= 120 nm

## Q1.6

First calculate the energy of each photon:

Since $$c=\lambda \nu$$, so $$\nu=\dfrac{c}{\lambda}$$, and the energy of each photon is:

$E_{ph}=h\nu=h\dfrac{c}{\lambda}=6.63\times 10^{-34} J s \times \dfrac{3.0 \times 10^{8} m/s}{1.06 \times 10^{4} nm \times \dfrac{1 \times 10^{-9} m}{1 nm}}=1.876 \times 10^{-20} J$

Second, calculate the energy needed to heat up 500 g of water by 10 degrees celsius:

$\Delta E_{water}=cm\Delta t=4.184 \dfrac{J}{g^oC}\times 500 g \times 10^oC=2.092\times 10^4 J$

Finally, calculate how many photons need to be absorbed to raise the temperature:

$number of photons=\dfrac{\Delta E_{Water}}{E_{ph}}=\dfrac{2.092\times 10^4 J}{1.876 \times 10^{-20} J}=1.115 \times 10^{24}$

## Q1.7

$C_{fusion}m_{ice}= (number of photons) \times E_{ph}= (number of photons) \times \dfrac{hc}{\lambda}$

So

$m_{ice}= (number of photons) \dfrac{hc}{\lambda}\dfrac{1}{C_{fusion}}=\dfrac{4.0\times 10^{23} \times 6.63\times 10^{-34} J s \times 3\times 10^8 m/s}{150 \times 10^{-9} m \times 333.55 \times 10^3 J/kg}=1.59 kg=1590 g$

Next calculate the mass of each water molecule:

The molar mass of water is $$MM_{water}=18 g/mol$$ and 1 mol is $$N_A$$ molecules, so the mass of each water molecule is

$m_{water molecule}=\dfrac{18 g}{N_A}=\dfrac{18 g}{6.022 \times 10^{23}}=2.989 \times 10^{-23} g$

So the number of water molecules that melted is:

$\dfrac{m_{ice}}{m_{water molecule}}=\dfrac{1590 g}{2.989 \times 10^{-23} g}=5.32\times 10^{25}$

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