1.

When the minimum wavelength of X-rays is 2Å then the applied potential difference between cathode and anticathode will be

Answer»

6.2 kV
2.48kV
24.8kV
62kV

Solution :`V = (12400)/(lamda_("min") (Å)) "VOLT " RARR V = 12400/2 = 6.2 KV`


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