Question

用波長為4×10^{－7 m}的紫光照射某金屬，發(fā)出的光電子垂直進(jìn)入3×10^－4 T的勻強(qiáng)磁場中，光電子所形成的圓軌道的最大半徑為1.2 cm.(電子電荷量e＝1.6×10^{－19 C}，質(zhì)量m＝0.91×10^{－30 kg)}求：

(1)光電子的最大初動能；

(2)該金屬發(fā)生光電效應(yīng)的極限頻率.

Answer 1

(1)1.82×10^－19 J  (2)4.74×10¹⁴ Hz

解析:

(1)由ev_mB=m                                                                                    (2分)

得  v_m=                                                                                                (2分)

則最大初動能為

E _k=mv_m²==1.82×10^－19 J.                                                              (2分)

(2)由E _k=h－W                                                                                            (2分)

W=hν₀                                                                                                                                                                                                                       (1分)

得極限頻率

ν₀==4.74×10¹⁴ Hz.                                                                           (1分)