Mookie Betts got off to a slow start to his 2019 campaign. But over the course of the last month, he has been red-hot. And during Thursday night's game against the Tampa Bay Rays, Betts quietly reached an important milestone.
Betts scored his 100th run of the season on Xander Bogaerts' first-inning homer. According to the MLB Stats Twitter account, Betts is the fastest Red Sox to reach that mark since Ted Williams in 1949.
And it turns out Betts is now in some elite company in Red Sox history.
Any time you're being compared to Baseball Hall of Famers like Ted Williams and Jimmie Foxx, it's a good thing. And it just demonstrates how good Betts has been at getting on base and putting himself in position to score.
Of course, it also helps Betts' cause that the trio of hitters behind him -- Rafael Devers, Xander Bogaerts, and J.D. Martinez -- have been three of the Red Sox best hitters this year and have done everything they can to drive in the leadoff hitter.
Boston Red Sox
That also explains why Devers (89) and Bogaerts (85) ranked second and third league-wide in runs scored heading into Thursday night's action, trailing only Betts. The top of the lineup has just been that good.
Betts didn't stay at the 100 run mark for very long. He hit a solo homer around the Pesky Pole in the seventh inning to plate his 101st run and cut the Rays' lead to three.
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