Question

我国发射的第一颗人造卫星采用20.009MHz和19.995MHz的频率发送无线电信号，这两种频率的电磁波的波长各是多少

Answer 1

20.009MHZ=2.0009×10 ⁷HZ，19.995MHZ=1.9995×10 ⁷HZ

λ ₁=

c

f 1 =

3× 10 8 m/s

2.0009× 10 7 HZ =14.993m

λ 2 =

c

f 2 =

3× 10 8 m/s

1.9995× 10 7 HZ =15.004m

答：这两种频率的电磁波的波长各是14.993m，15.004m．