作业帮求解几道英文化学题1）The binding energy of electrons to copper metal i
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求解几道英文化学题
1）The binding energy of electrons to
copper metal is 383.0 kJ/mol. What is the longest wavelength of light that will
eject electrons from copper metal? Enter units
2）Microwave ovens heat food by creating microwave electromagnetic radiation that
is absorbed by water molecules in the food. Any material that does not have
water in it will not absorb the radiation and will not get hot. Metals reflect
the microwaves from their surfaces and disrupt the operation of the oven. Some
Australians heat water in microwave ovens to make tea. What is the minimum
number of microwave photons with a wavelength of 3.22 mm that will have to be
absorbed by a 122.00 gram sample of water to heat it from 25°C to 100°C? (It
requires about 315 joules of energy to heat 1 gram of water 75°C)
3）What is the wavelength (in nm) of an electron with v = 9.2x 105
m/s?
4）Calculate λ (in nm) for a photon with E = 2.83x10-15 J?
5）Give the number of unpaired electrons in F−.
1）The binding energy of electrons to
copper metal is 383.0 kJ/mol. What is the longest wavelength of light that will
eject electrons from copper metal? Enter units
2）Microwave ovens heat food by creating microwave electromagnetic radiation that
is absorbed by water molecules in the food. Any material that does not have
water in it will not absorb the radiation and will not get hot. Metals reflect
the microwaves from their surfaces and disrupt the operation of the oven. Some
Australians heat water in microwave ovens to make tea. What is the minimum
number of microwave photons with a wavelength of 3.22 mm that will have to be
absorbed by a 122.00 gram sample of water to heat it from 25°C to 100°C? (It
requires about 315 joules of energy to heat 1 gram of water 75°C)
3）What is the wavelength (in nm) of an electron with v = 9.2x 105
m/s?
4）Calculate λ (in nm) for a photon with E = 2.83x10-15 J?
5）Give the number of unpaired electrons in F−.
1）hc/λ＞383×10^3/NA
h：普朗克常数6.63×10^-34J·s,c：光速3×10^8m/s,λ：光子波长,NA：阿伏伽德罗常数6.02×10^23
解得波长＜3.13×10^-7m,即313nm
2）意思就是微波加热需要有水分子,如果没有水而是金属的话会反射微波损坏微波炉.
1个3.22mm微波光子能量E=hc/λ=6.18×10^-23J
122g水从25℃加热到100℃需要热量为122*315=38430J
需要的微波光子的数量最少为：38430/6.18×10^-23=6.22×10^26个
3）根据德布罗意波长公式λ=h/mv=6.63×10^-34/(9.11×10^-31*9.2×10^5)=7.94×10^-10m=0.794nm
公式中m：电子质量9.11×10^-31kg,v：电子运动速率,h：普朗克常数
4）根据爱因斯坦光子理论E=hc/λ,算出λ=7.03×10^-11m=0.07nm
5）F-（是氟离子吧）未成对电子数为0,就是说所有电子都是成对的.
h：普朗克常数6.63×10^-34J·s,c：光速3×10^8m/s,λ：光子波长,NA：阿伏伽德罗常数6.02×10^23
解得波长＜3.13×10^-7m,即313nm
2）意思就是微波加热需要有水分子,如果没有水而是金属的话会反射微波损坏微波炉.
1个3.22mm微波光子能量E=hc/λ=6.18×10^-23J
122g水从25℃加热到100℃需要热量为122*315=38430J
需要的微波光子的数量最少为：38430/6.18×10^-23=6.22×10^26个
3）根据德布罗意波长公式λ=h/mv=6.63×10^-34/(9.11×10^-31*9.2×10^5)=7.94×10^-10m=0.794nm
公式中m：电子质量9.11×10^-31kg,v：电子运动速率,h：普朗克常数
4）根据爱因斯坦光子理论E=hc/λ,算出λ=7.03×10^-11m=0.07nm
5）F-（是氟离子吧）未成对电子数为0,就是说所有电子都是成对的.
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