物理計算題

2009-06-12 6:46 pm
1.在真空中若一平面電磁波之頻率為3x10^9(Hz),則其波長為_____m。若其最大電場強度為60Nt/C,則其最大磁場強度為 ___________ Nt/A-m。
2. 波長為150nm的光其每一個光子所帶的能量是 _____ J。把一個電子從金屬鋁中移出需要能量4.2Ev,若將波長為150nm的光照在金屬鋁上,從金屬鋁中跑出的電子所具有的最大動能為 ________ J。

可以解釋一下第一條第二個小問是用什麼公式嗎?

回答 (1)

2009-06-13 8:23 pm
✔ 最佳答案
1. wavelength = speed of wave in vacuum/frequency of wave
=3 x 10^8/3 x 10^9 m = 0.1 m
where 3 x 10^8 m/s is the speed of eelctromagentic wave in free space.
Magnetic flux density = electric field intensity/speed of light
= 60/3x10^8 N/A-m = 2 x 10^-7 N/A-m
2. 150 nm = 150 x 10^-9 m = 1.5 x 10^-7 m
Energy of photon = hc/w
where h is Planck's constant (= 6.6 x 10^-34 Js)
c is the speed of light (= 3 x 10^8 m/s)
w is the wavelength of light (= 1.5x10^-7 m)
hence, energy of photon = 1.32 x 10^-18 J
Since 4.2 eV = 4.2 x (1.6x10^-19) J = 6.72 x 10^-19 J
[where 1.6x10^-19 coulomb is the electronic charge]
hence, kinetic energy of electron
= (1.32 x 10^-18 - 6.72 x 10^-19) J = 6.48 x 10^-19 J
可以解釋一下第一條第二個小問是用什麼公式嗎?
The formula used is: E = c.B
where E is the eelctric field intensity (in unit of v/m, or N/C)
B is the magnetic flux density (in units of N/A-m, or wb/m2, or T)
c is the speed of light in vacuum (in unit of m/s)


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