Respuesta :

f = W/h = (1.98*1.6*10^-19)/(6.63*10^-34) = 4.78*10^14 Hz  

The threshold frequency in the metal is 4.8×10⁻¹⁴hz

The first step is to write out the parameters given in the question

The work function of the metal is 1.98 ev

Planck function is 6.6 ×10⁻34

Threshold function(v) = ?

The threshold frequency can be calculated as follows;

v= 1.98×1.6×10⁻¹⁹/6.6×10⁻³⁴

= 3.168×10⁻¹⁹/6.6×10⁻³⁴

= 4.8×10¹⁴

Hence the threshold frequency in the metal is 4.8×10⁻¹⁴ hz

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